IN   MEMORIAM 
FLOR1AN  CAJORI 


'a*    (St^isrc 


<K 


*\ 


■<->.►■.  o*» 


HIGHER    ALGEBRA 


BY 


JOHN   F.   DOWNEY,   M.A.,  C.E. 

PROFESSOR   OF   MATHEMATICS   IN   THE   UNIVERSITY   OF   MINNESOTA 


?XKc 


NEW   YORK-:.  CINCINNATI. -CHICAGO 

AMERICAN     BOOK    COMPANY 


Copyright,  1900,  by 
JOHN  F.   DOWNEY. 


^DOWNEY'S.  HIGH.  ^ALG. 


CAJORI 


PREFACE 


This  work  is  designed  as  a  text-book  in  universities,  colleges, 
and  technical  schools,  the  first  fifteen  chapters  being  also' adapted 
to  use  in  high  schools  and  academies  by  students  who  have  some 
knowledge  of  elementary  algebra. 

The  demonstrations  constitute  one  of  the  characteristic  features 
of  the  book.  While  most  of  our  text-books  on  Algebra  state  with 
great  clearness  the  theorems  and  rules,  few  of  them,  especially 
in  the  earlier  parts,  give  the  demonstrations  in  a  way  that  enables 
a  student  to  reproduce  them.  Usually  illustration,  explanation, 
and  general  demonstration  are  so  intermingled  that  the  student 
is  not  able  to  gather  up  and  give  in  logical  form  just  what  con- 
stitutes the  proof.  In  this  work  the  plan  is  that  which  gives  so 
much  definiteness  to  our  teaching  in  Geometry:  each  general 
principle  is  followed  by  a  concise,  logical  demonstration,  contain- 
ing only  the  reasoning  necessary  to  establish  it,  while  all  illus- 
trations and  explanations  by  special  cases  are  given  in  separate 
articles.  The  student  thus  soon  learns  to  know  what  is  demanded 
in  a  general  proof,  and  to  distinguish  between  rigorous  demon- 
stration and  verification  or  illustration  by  a  special  case.  With- 
out any  loss  of  conclusiveness  in  reasoning,  the  methods  employed 
have  permitted,  in  many  cases,  much  shorter  and  more  easily 
followed  demonstrations  than  those  usually  given. 

Another  characteristic  feature  is  the  substitution  of  short 
processes  for  many  of  the  long  and  tedious  ones  in  common  use. 
As  mathematical  operations,  at  best,  involve  much  drudgery,  all 
practical  means  of  shortening  the  work  should  be  made  available 
to  the  student.  The  few  short  processes  given  in  our  text-books 
are  reserved  until  the  student  has  formed  a  habit  of  using  the 
long  processes,  and,  consequently,  he  never  gains  a  practical  use 
of  even  these  few.     In  this  book  short  processes  are  introduced 

3 


911285 


4  PREFACE 

at  the  beginning  of  the  respective  subjects,  and  are  used  wherever 
applicable.  For  example,  a  very  expeditious  method  of  multiply- 
ing by  a  factor  of  the  form  x  ±  a  or  x  ±  ay  is  fully  explained, 
and  is  applied  to  successive  multiplication,  to  putting  together 
the  factors  that  constitute  the  highest  common  divisor  and  the 
lowest  common  multiple,  to  forming  an  equation  having  given 
roots,  and  to  various  other  operations.  Again,  the  equally  expe- 
ditious method  of  dividing  by  a  factor  of  the  form  x  ±  a  or 
x  ±  ay,  instead  of  being  confined  to  finding  the  commensurable 
roots  of  higher  equations,  is  fully  explained  in  Division,  and  is 
applied  to  successive  division,  to  factoring  polynomials,  to  find- 
ing the  highest  common  divisor  and  the  lowest  common  multiple, 
to  reducing  fractions  to  their  lowest  terms,  to  finding  the  com- 
mensurable roots  of  higher  equations,  and  to  various  other  opera- 
tions. The  student  thus  becomes  very  familiar  with  the  processes, 
and  expert  in  their  use;  and  he  not  only  escapes  much  tedious 
labor  that  has  no  disciplinary  value,  but  secures  greater  accuracy 
in  his  results,  the  shorter  processes  diminishing  the  liability  to 
errors. 

The  methods  given  of  obtaining  at  once  the  square  root  and 
the  cube  root  of  polynomials,  without  writing  any  intermediate 
steps,  will  be  found  not  only  a  great  relief  from  the  tedious 
processes  given  in  our  books,  but  an  exhilarating  exercise  in 
which  students  will  take  great  interest. 

The  subject  of  Maxima  and  Minima  of  Functions  is  presented 
in  a  fuller  and  more  systematic  way  than  heretofore,  and  the 
application  made  to  practical  problems  cannot  fail  to  interest  the 
student. 

While  those  properties  of  higher  equations  which  serve  no 
useful  purpose  have  been  excluded,  the  subject  has  received  a 
fuller  treatment  than  is  common.  The  reader  will  find  many 
features  presented  in  new  and  simpler  ways,  with  everything 
leading  toward  the  easiest  and  most  expeditious  methods  of 
finding  the  roots  of  numerical  higher  equations.  The  process  of 
finding  the  roots  of  equations  having  only  even  or  only  odd 
powers  will  be  seen  to  be% remarkably  brief. 

Differentiation  of  algebraic  and  logarithmic  functions  is  intro- 


PREFACE  5 

duced,  because  it  enables  us  to  give  in  their  true  relations  and 
with  their  proper  significance  differential  coefficients,  which  are 
usually  disguised  under  the  name  of  "derived  polynomials"; 
because  it  enables  us  to  include  Taylor's  Formula,  and  thus  avoid 
the  usual  cumbersome  and  unsatisfactory  method  of  demonstrat- 
ing the  binomial  formula  and  the  logarithmic  series ;  because  it 
puts  into  its  true  relation  f'(x)  in  the  theorem  for  multiple  roots 
and  in  Sturm's  Theorem;  and  because  in  his  subsequent  course 
in  mathematics  the  student  will  not  use  the  methods  which  dif- 
ferentiation avoids,  and  will  constantly  use  the  methods  which 
it  introduces.  Differentiation  is  given  by  the  fluxionary  system, 
as  furnishing  the  clearest  and  most  satisfactory  conceptions  and 
the  simplest  demonstrations. 

Many  exercises  and  problems  have  been  given  in  all  the  dif- 
ferent subjects.  These  will  be  found  well  graded,  and  neither 
so  difficult  as  to  discourage  the  student  nor  so  easy  as  to  afford 
little  discipline.  To  guide  the  student  in  the  application  of  prin- 
ciples to  numerical  examples,  such  suggestions,  observations,  and 
model  solutions  as  long  experience  in  teaching  the  subject  has 
shown  to  be  needful  have  been  given.  While  many  of  the  exer- 
cises and  problems  are  new,  free  use  has  been  made  of  the  various 
collections  available. 

The  subject  of  Determinants  is  not  included,  because  inexpen- 
sive texts  devoted  exclusively  to  the  subject  can  be  readily  pro- 
cured by  the  comparatively  small  number  of  students  who  have 
occasion  to  enter  upon  the  subject  at  this  stage  of  their  progress. 
The  chapter  on  the  Theory  of  Functions  will  be  found  of  more 
value  to  the  student  than  would  be  an  elementary  treatment  of  the 
Loci  of  Equations,  as  it  gives  him  needed  practice  in  the  inter- 
pretation of  analytical  results,  and  thus  prepares  him  for  the 
subject  of  Loci  when  he  reaches  it  in  Analytical  Geometry. 

Assistance  has  been  received  from  so  many  sources  that  no 
attempt  is  made  to  name  them.  While  in  many  respects  the 
book  is  a  wide  departure  from  the  texts  of  the  day,  nothing  has 
been  made  different  for  the  sake  of  novelty.  WThether  new  or 
old,  the  methods  which  long  experience  with  large  classes  has 
proved  to  be  the  best  have  been  given. 


6  PREFACE 

The  book  is  put  forth  with  the  hope  that  it  will  not  only  serve 
well  the  usual  purpose  of  a  text  on  Algebra,  but  that  it  will,  in 
addition,  cultivate  in  students  much  greater  facility  in  the  expla- 
nation of  principles,  and  that  it  will  relieve  them  from  a  great 
burden  of  worse  than  needless  details  in  operations,  which  many 
generations  of  students  have  been  required  to  carry. 


CONTENTS 


CHAPTER  PAGE 

I.     Definitions  and  Notation 15 

Quantity 15 

Algebra 15 

The  Symbols  of  Quantity 16 

The  Symbols  of  Operation 17 

The  Symbols  of  Relation 18 

The  Symbols  of  Aggregation 19 

The  Symbols  of  Abbreviation 19 

Algebraic  Expressions 19 

Positive  and  Negative  Quantities 20 

Axioms,  Theorems,  etc. 20 

II.     Addition 22 

Addition  of  Monomials 22 

Addition  of  Polynomials 23 

Addition  of  Partially  Similar  Terms 25 

Addition  of  Compound  Terms 25 

III.  Subtraction 26 

Rule  for  Subtraction 27 

Signs  of  Aggregation  as  related  to  Addition  and  Subtraction  28 

IV.  Multiplication 30 

The  Sign  of  the  Product 31 

Multiplication  of  Monomials 31 

Multiplication  of  Polynomials 32 

Homogeneous  Polynomials 33 

Multiplication  by  Detached  Coefficients         ....  31 

Short  Methods  of  Multiplication 36 

By  a  factor  of  the  form  x  ±  a 36 

Successive  Multiplication 39 

When  the  factors  are  of  the  form  x  ±  a  and  x±  b      .        .  40 

The  Square  of  the  Sum  and  of  the  Difference     ...  42 

The  Product  of  the  Sum  and  Difference      ....  42 

The  Square  of  Any  Polynomial 43 

The  Sum  of  Two  Groups  by  the  Difference        ...  43 
7 


8  CONTENTS 

PAGE 

V.     Division 45 

The  Signs  of  the  Quotient 45 

Division  of  Monomials 45 

Exponent  0 45 

Transference  of  Factors  from  Dividend  to  Divisor,  and  vice 

versa 46 

Division  of  Polynomials 47 

Short  Methods  of  Division 49 

Horner's  Synthetic  Division .50 

When  the  divisor  is  of  the  form  x  ±  a        .         .         .         .53 

Successive  Division 57 

When  the  divisor  is  separable  into  factors  of  the  form 

x  ±  a 58 

VI.     Factoring 59 

Monomial  Factors 59 

Trinomial  Squares 59 

Difference  of  Two  Squares 59 

Six-term  Squares 59 

Trinomials  of  the  Form  x2m  +  pxm  +  q 61 

Trinomials  of  the  Form  ax2m  +  bxm  -f  c  .         .         .         .62 

Separating  into  Groups 63 

The  Difference  of  the  Same  Powers 65 

The  Sum  of  the  Same  Powers        .         .  ■      .         .  .65 

Binomial  Factors  by  Trial 66 

VII.     Highest  Common  Divisor  and  Lowest  Common  Multiple   .  71 

Highest  Common  Divisor 71 

Polynomials  readily  factored 71 

One  of  Two  Polynomials  readily  factored  ....  74 

Polynomials  not  readily  factored 75 

Lowest  Common  Multiple 77 

VIII.     Fractions 79 

Keduction  of  Fractions 80 

To  Lowest  Terms 81 

To  Integral  or  Mixed  Forms        .        .        .         .         .         .82 

Mixed  Quantity  to  Fractional  Form 83 

To  Lowest  Common  Denominator 84 

Addition  and  Subtraction  of  Fractions 85 

Multiplication  of  Fractions 86 

A  Fraction  by  an  Integer 86 

A  Fraction  by  a  Fraction 87 


CONTENTS 


IX. 


Division  of  Fractions     . 

A  Fraction  by  an  Integer 

A  Fraction  by  a  Fraction 
Simplification  of  Fractions 

Complex  Fractions 

Terminated  Continued  Fractions 

Fractions  with  Negative  Exponents 


Theory  of  Exponents,  Involution,  and  Evolution 
Theory  of  Exponents     .... 
Fractional  and  Negative  Exponents    . 

(abc  .  .  .)n  =  anbncn 

To  affect  a  Monomial  with  Any  Exponent 

Involution 

Monomials 

Binomials  —  Binomial  Theorem 
Cube  of  a  Polynomial 

Evolution 

Monomials 

Index  a  Composite  Number 

Square  Root  of  Polynomials  by  Inspection 

Trinomials 

Only  Two  Powers  of  One  Letter     . 

Any  Other  Perfect  Square 
Cube  Root  of  Polynomials  by  Inspection 

Quadrinomials         .... 

Only  Three  Powers  of  One  Letter  . 

Any  Other  Perfect  Cube 
Any  Root  of  Any  Quantity 


X. 


Surds  and  Imaginaries     . 
Surds      .... 

Reduction  of  Surds     . 
Removal  of  a  Factor 
Diminishing  the  Degree 
Rational  Quantity  to  Radical  Form 
Introduction  of  Coefficient 
To  Equivalent  Surds  of  Same  Degree 
Rationalizing  a  Denominator  . 
The  Rationalizing  Factor  of  a  Binomial 

Addition  and  Subtraction  of  Surds 

Multiplication  of  Surds 

Division  of  Surds        .... 


Surd 


10 


CONTENTS 


XI. 


XII. 


XIII. 


Involution  of  Surds     . 
Evolution  of  Surds 
Imaginary  Quantities 
General' Form      . 
The  Imaginary  Element 
Index  Composite 
Conjugate  Imaginaries 
Modulus  of  a  Quadratic  Imaginary 
Addition  and  Subtraction  of  Imaginaries 
The  Different  Powers  of  ^/~=T  and  ^/^H 
Multiplication  and  Division  of  Imaginaries 
Use  of  Imaginaries 


Simple  Equations 

One  Unknown  Quantity 
Axioms       .... 
Rule  for  Solution 
Some  Practical  Suggestions 
Equations  with  Surd  Terms 
Problems     .... 
Two  Unknown  Quantities 
Elimination 

By  Addition  or  Subtraction 
By  Comparison 
By  Substitution 
Some  Practical  Suggestions 
Problems 
Several  Unknown  Quantities 
Problems    .... 


Inequalities       .... 
Operations  that  do  not  change  the 
Operations  that  change  the  Sense 
Caution 


Sense 


Ratio,  Proportion,  and  Variation 

Ratio 

Operations  that  do  not  change  a  Ratio 
Examples  and  Problems 

Proportion 

An  Equation  from  a  Proportion 
A  Proportion  from  an  Equation 
Operations  that  do  not  destroy  a  Proportion 


CONTENTS  11 

PAGE 

Examples  and  Problems 184 

Variation 187 

Ratio  given  by  a  Set  of  Values 189 

From  a  Variation  to  an  Equation,  and  vice  versa      .         .190 

From  a  Variation  to  a  Proportion 190 

Examples  and  Problems 190 

XIV.     Progressions 195 

Arithmetical  Progression 195 

Formulae 196 

Examples  and  Problems 197 

Geometrical  Progression       .         .         .    .     .        .         .         .  200 

Formula 201 

Examples  and  Problems 203 

Harmonic  Progression 205 

XV.     Quadratic  Equations 208 

Pure  Quadratics 208 

Rule  for  solving 208 

Character  of  the  Roots 209 

Problems 209 

Affected  Quadratics 211 

Rule  for  solving 211 

Character  of  the  Roots      .         .         .         .  m       .         .         .212 

Problems 213 

Solution  by  Factoring 216 

Formation  of  an  Equation  having  Given  Roots         .         .217 

Introduction  and  Loss  of  Roots 218 

XVI.     Some  Higher  Equations 222 

Pure  Equations 222 

Equations  in  the  Quadratic  Form         .        ...        .        .  223 

Equations  with  Integral  Roots 226 

XVII.     Equations  of  the   Second   Degree  with   Two   Unknown 

Quantities 230 

One  of  the  Equations  of  the  First  Degree     ....  231 
All  Terms  but  One  each  of  First  Degree       .         .         .         .232 

One  of  the  Equations  Homogeneous 232 

Both  Homogeneous  except  in  Absolute  Term       .         .        .  233 

Symmetrical  Equations 238 

Problems 240 


12  CONTENTS 


PACE 

XVIII.     Theory  of  Functions 242 

Maxima  and  Minima  of  Functions        .  242 

Quadratic  Functions 245 

Reciprocals  of  Quadratic  Functions 247 

Algebraic  Functions  of  Any  Form 248 

Problems 252 

Zero,  Infinity,  and  Indeterminate  Forms     ....  255 

The  Forms  -£  and  ^ 255 

0  co 

The  Forms  -  and  — 255 

0  GO 

Discussion  of  Functions  and  Problems         ....  257 

Interpretation  of  Negative  Results 257 

Interpretation  of  Imaginary  Results          ....  258 

Examples  and  Problems 259 

XIX.     Differentiation  of  Algebraic  Functions  ....  263 

Constant  Terms 263 

The  Product  of  a  Constant  and  a  Variable  .        .  *     .        .  263 

A  Polynomial '  .         .  264 

The  Product  of  Two  Variables 264 

The  Product  of  Several  Variables 265 

A  Fraction 265 

A  Variable  having  a  Constant  Exponent      ....  266 

Second,  Third,  etc.,  Differential 268 

First,  Second,  Third,  etc.,  Derivative  .....  268 

Partial  Derivatives .270 

XX.     Development  of  Functions 272 

Indeterminate  Coefficients 274 

The  Law  of  the  Series 275 

Decomposition  of  Fractions 278 

Taylor's  Formula 282 

The  Binomial  Formula 285 

XXI.     Logarithms 288 

Logarithm  of  a  Product 289 

Logarithm  of  a  Quotient       .......  290 

Logarithm  of  a  Power 290 

Logarithm  of  a  Root .  290 

Law  of  Characteristics 291 

The  Mantissa  and  the  Decimal  Point 292 

The  Differential  of  a  Logarithm 292 

The  Logarithmic  Series ,  293 


CONTENTS 


13 


PAGE 

Relation  of  Logarithms  in  Different  Systems       .        .        .  294 

The  Logarithmic  Series  rendered  Convergent       .         .         .  295 

Computation  of  Logarithms 295 

Tables  of  Logarithms    .        .         .        ••■'-.         .        .        .  296 

Values  of  m  and  e         .......  296 

Exponential  Equations         .......  297 

Compound  Interest  by  Logarithms       ,        .        .        .        .  297 

Present  Worth  of  an  Annuity 298 

Explanation  of  Tables 299 


XXII.     Indeterminate  Equations 
Problems 


305 
309 


XXIII.     Theory  of  Equations  and  Solution  of  Numerical  Higher 
Equations   . 
Reduction  to  the  Normal  Form    . 
Test  for  Roots      .... 

Short  Method  of  Substitution    . 

Test  for  Roots  by  Division 
Number  and  Character  of  the  Roots 

Normal  Equation  has  no  Fractional  Roots 

An  Equation  has  n  Roots 

Formation  of  an  Equation  from  Roots 

Imaginary  Roots  in  Conjugate  Pairs 

Negative  Roots 

Descartes's  Rule  of  Signs 
Solution  for  Commensurable  Roots 
Equations  with  only  Even  or  only  Odd  Powers 
Relation  of  Roots  to  Coefficients  . 
Situation  of  Roots 
Multiple  Roots      .... 

Method  by  H.  C.  D.  of /(x)  and /'(a) 

Better  Methods 
Solution  for  Incommensurable  Roots 

Horner's  Method 

Diminution  of  Roots  by  a  Given  Quantity 

Approximate  Value  of  a  Fractional  Root 

Principle  of  Horner's  Method    . 
Observations  on  Horner's  Method 
Examples  and  Problems 
The  nth  Root  of  a  Number    . 

Sturm's  Theorem  and  Method  . 
The  Theorem  for  Situation  of  Roots 


310 
311 
313 
314 
314 
315 
315 
316 
316 
316 
317 
318 
320 
322 
323 
325 
328 
328 
329 
331 
331 
331 
334 
335 
336 
337 
339 
340 
341 


14  CONTENTS 

PAGE 

Recurring  or  Reciprocal  Equations 347 

Binomial  Equations 350 

Cardan's  Solution  of  the  General  Cubic  Equation        .         .  351 

Descartes's  Solution  of  the  General  Biquadratic  Equation  .  353 

XXIV.     Series 354 

Convergency  of  Series 354 

Scale  of  Relation  of  a  Recurring  Series        ....  360 

The  nth  Term  of  a  Series 363 

Summation  of  a  Series          .......  365 

By  the  Method  of  Decomposition 366 

By  the  Scale  of  Relation 370 

By  the  Method  of  Differences 372 

Piles  of  Spherical  Shot 373 

Interpolation 375 

XXV.     Permutations  and  Combinations 379 

Permutations 379 

Problems 381 

Combinations .  383 

Problems 387 

Answers          .                 389 


HIGHER  ALGEBRA 


CHAPTER   I 
DEFINITIONS   AND  NOTATION 

1.  Quantity  is  amount  or  extent,  and  is  expressed  in  terms  of  a 
unit  of  the  same  kind. 

Thus,  10,  5  bushels,  6  tons,  50  miles,  7  years,  100  dollars,  m  square  feet, 
n  cubic  yards,  are  quantities,  the  units  in  order  being  the  abstract  number  1, 
1  bushel,  1  ton,  1  mile,  1  year,  1  dollar,  1  square  foot,  1  cubic  yard. 

2.  Two  or  more  quantities  of  the  same  kind  are  Commensurable 
or  Incommensurable  with  reference  to  one  another  according  as 
they  can  or  can  not  be  measured  with  the  same  unit. 

Thus,  if  the  sides  of  a  rectangle  are  3  feet  and  4  feet  respectively,  the 
diagonal  is  5  feet,  and  is,  therefore,  commensurable  with  the  sides ;  but  the 
diagonal  of  a  square  is  incommensurable  with  its  sides,  being  V2  times  one 
of  the  sides. 

3.  An  Incommensurable  Quantity,  without  comparison  with  an- 
other quantity,  is  one  that  cannot  be  exactly  expressed  in  the 
decimal  notation. 

Thus,  V5  is  an  incommensurable  quantity,  being  2  plus  a  decimal  fraction 
which  never  terminates. 

4.  Algebra  is  that  branch  of  pure  mathematics  which  treats  of 
numbers  as  expressed  by  symbols  having  general  values,  and  of 
the  nature,  transformations,  and  use  of  equations. 

5.  Algebra  differs  from  arithmetic  in  the  following  important 
particulars : 

15 


}'o  HIGHER  ALGEBRA 

1st.  In  arithmetic  values  are  counted  in  only  one  direction 
from  0;  while  in  algebra,  by  means  of  positive  and  negative 
quantities,  values  are  counted  in  two  opposite  directions  from  0. 

2d.  Arithmetical  quantities,  denoted  by  figures,  have  each  a 
single,  definite  value ;  while  algebraical  quantities,  represented  by 
letters,  may  have  any  value  we  choose  to  assign  to.  them.  These 
quantities  can  be  recognized  anywhere  in  the  operation,  and  the 
results  are,  therefore,  general  formulae  instead  of  special  answers. 

3d.  In  arithmetic  a  problem  is  solved  by  analyzing  it,  step  by 
step ;  while  in  algebra  the  conditions  are  expressed  in  equations 
involving  one  or  more  unknown  quantities,  usually  representing 
the  answer  or  answers. 

6.  Algebra  employs  five  kinds  of  symbols,  viz.  of  quantity,  of 
operation,  of  relation,  of  aggregation,  and  of  abbreviation. 

7.  The  Symbols  of  Quantity  commonly  employed  are  the  fol- 
lowing : 

1st.   The  Arabic  figures. 

2d.  The  letters  of  the  Roman  alphabet,  known  quantities  being 
represented  by  the  leading  letters,  and  unknown  quantities  by  the 
final  letters. 

Similar  quantities  employed  in  the  demonstration  of  a  theorem 
or  the  solution  of  a  problem  are  often  represented  by  the  same 
letter  with  different  accents,  as  a',  a",  a'",  etc.,  read  ua  prime," 
"  a  second,"  "  a  third,"  etc.,  or  by  the  same  letter  with  different 
subscripts,  as  alf  a2,  a3,  etc.,  read  "  a  sub-one,"  "  a  sub-two,"  "  a 
sub-three,"  etc. 

Initial  letters  are  sometimes  used,  as  r  for  radius,  s  for  sum,  d 
for  difference,  etc. 

The  Greek  -*  is  used  for  the  ratio  of  the  circumference  of  a 
circle  to  its  diameter. 

3d.  Zero,  0,  used  to  denote  not  only  the  absence  of  value,  but 
also  a  quantity  that  is  less  than  any  assignable  value. 

4th.  Infinity,  oo,  used  to  represent  a  quantity  that  is  greater 
than  any  assignable  value. 


DEFINITIONS  AND  NOTATION  17 

8.  The  Symbols  of  Operation  in  algebra  are  those  common  to  all 
branches  of  mathematics,  the  principal  ones  being  the  following : 

1st.   The  sign  of  addition,  +  ,  read  "plus." 
2d.    The  sign  of  subtraction,  — ,  read  "minus." 

3d.  The  sign  of  multiplication,  x,  read  "times,"  "into,"  or 
"  multiplied  by." 

Multiplication  is  indicated  also  by  a  dot  between  the  quantities. 

The  quantities  between  which  multiplication  is  indicated  are 
called  Factors,  and  the  result  of  the  multiplication  is  called  the 
Product. 

4th.   The  sign  of  division,  ■+-,  read  "divided  by." 
Division  is  indicated  also  by  writing  the  dividend  above,  and 
the  divisor  below,  a  line,  in  the  fractional  form. 

5th.  The  sign  of  evolution,  -y/,  called  the  radical  sign,  and  read 
"  the  square  root  of,"  the  square  root,  or  second  root,  being  one  of 
the  two  equal  factors  into  which  a  number  is  conceived  to  be 
resolved.  The  3d,  4th,  or  nth  root,  by  which  is  meant  one  of  the 
3,  4,  or  n  equal  factors  into  which  a  number  is  conceived  to  be 
resolved,  is  indicated  by  writing  3,  4,  or  n  in  the  vertex  of  the 
angle  of  the  sign,  this  number  being  called  the  Index. 

9.  Besides  the  above  signs,  or  symbols  of  operation,  there  are 
certain  positions  of  quantities  with  reference  to  other  quantities 
that  indicate  operations,  the  principal  ones  being  the  following : 

1st.  A  Coefficient,  which  is  a  quantity  written  beside  another 
quantity  to  show  how  many  times  the  latter  is  taken. 

When  algebraic  quantities  are  written  in  succession  with  no 
sign  between  them,  their  product  is  signified,  and  any  factor,  or 
the  product  of  any  number  of  factors,  is  the  coefficient  of  the 
product  of  the  remaining  factors. 

Thus,  in  5  mnx,  5  is  the  coefficient  of  mnx,  5  m  is  the  coefficient  of  nx, 
and  5  mn  is  the  coefficient  of  x. 

The  name  is  usually  applied  to  the  numerical  factor,  and  when 
this  is  1,  it  is  not  written. 

DOWNEY'S    ALG.  — 2 


18  HIGHER  ALGEBRA 

2d.  An  Exponent,  which  is  a  quantity  written  at  the  right  of 
and  above  another  quantity,  its  meaning  being  as  follows : 

(a)  When  an  exponent  is  a  positive  integer,  it  indicates  that  the 
quantity  affected  by  it  is  to  be  taken  as  a  factor  as  many  times 
as  there  are  units  in  the  exponent.  The  result  of  the  multiplica- 
tion is  called  a  Power ;  hence  a  positive  integral  exponent  is  said 
to  indicate  a  power. 

Thus,  a  •  a  =  a2,  read  "  a  square,"  "  a  2d  power,"  or  "  a  2d  "  ; 

a  •  a  •  a  =  a3,  read  "a  cube,"  "a  3d  power,"  or  "a  3d"  ; 
a  •  a  •  a  •  a  =  a4,  read  "  a  4th  power,"  or  "  a  4th  "  ; 
a  •  a  •  a  •••  to  n  factors  =  an,  read  "  a  nth  power,"  or  "  a  nth." 

(b)  When  an  exponent  is  a  positive  fraction,*  the  numerator 
indicates  a  power  and  the  denominator  a  root  of  the  quantity 
affected  by  it. 

Thus,  32^  is  the  4th  power  of  the  5th  root  of  32,  or  the  5th  root  of  the  4th 
power  of  32. 

(c)  When  an  exponent  is  negative,  it  indicates  the  reciprocal  of 
what  it  would  indicate  if  it  were  positive,  the  reciprocal  of  a 
quantity  being  1  divided  by  that  quantity. 

Thus,  a-2  =  1  and  a2&"3  =  -• 

a2  W 

10.   The  Symbols  of  Relation  are  the  following : 

1st.    The  sign  of  equality,  =,  read  "equals,"  or  "is  equal  to." 

2d.  The  signs  of  inequality,  >  and  <,  read  respectively,  "is 
greater  than,"  and  "  is  less  than." 

The  signs,  =£,  >,  and  <,  read  respectively,  "is  not  equal  to," 
"is  not  greater  than,"  and  "is  not  less  than,"  are  also  employed 
to  some  extent. 

3d.   The  sign  of  geometrical  ratio,  : ,  read  "  to,"  or  "  is  to." 

*  It  will  be  shown  in  Art.  147  that  if  we  assume  the  index  law  as  proved 
for  positive  integral  exponents  to  be  general,  the  meaning  of  positive  frac- 
tional exponents  and  negative  exponents  must  be  as  defined  above.  Until 
that  article  is  reached,  their  meaning  will  be  treated  as  a  matter  of  definition. 


DEFINITIONS  AND  NOTATION  19 

4th.   The  sign  of  equality  of  ratios,  : :,  read  "as,"  or  "equals." 

Thus,  a:b  :tc:d  is  read  either  "  a  is  to  6  as  c  is  to  d,"  or  "  the  ratio  of 
a  to  b  equals  the  ratio  of  c  to  tV 

5th.   The  sign  of  variation,  oc,  read  "  varies  as." 

11.   The  Symbols  of  Aggregation  are  the  parentheses  (  ),  the 

brackets  [  ],  the   braces    \  j,  and  the  vinculum  ,  and  all 

indicate  that  the  algebraic  expression  included  is  to  be  treated 
as  a  whole. 


Thus,  a  —  (o  +  c),  a  —  [6  +  c],  a  —  {b  +  c},  and  a  —  b  +  c  all  indicate 
that  the  sum  of  b  and  c  is  to  be  subtracted  from  a. 

The  vinculum  is  sometimes  vertical. 


Thus,  a 
b 
c 


x  is  the  same  as  a  +  b  +  c  x  x. 


The  line  between  the  numerator  and  denominator  of  a  fraction 
has  the  effect  of  a  vinculum. 

b  -4-  c 
Thus,  a -£—  is  the  same  as  a  —  $(b  +  c). 


12.  The  Symbols  of  Abbreviation  are  the  following: 

1st.   The  signs  of  deduction,  .*.  and  v,  read  respectively,  "there- 
fore "  or  "  hence,"  and  "  since  "  or  "  because." 

2d.   The  sign  of  continuation,  •••,  read  "and  so  on,"  or  "and  so 
on  to." 

Thus,  1  +  3  +  5  +  7  +  -. -is  read  "  1  plus  3  plus  5  plus  7  and  so  on,"  and 
a  +  ar  +  ar2  +  •••  arn~*  is  read  M  a  plus  ar  plus  ar2  and  so  on  to  ar"-1.'* 

3d.   The  factorial  sign,  L,  which  indicates  the  product  of  all  the 
integers  from  1  to  the  number  written  in  the  angle  inclusive. 

Thus,  [8  means  2  x  3,  [_5  means  2  x  3  x  4  x  5,  [n  means  2  x  3  x  4  •••  n. 

13.  An  Algebraic  Expression  is  a  quantity,  simple  or  made  up 
of  parts,  expressed  in  algebraic  symbols. 


20  HIGHER  ALGEBRA 

14.  The  Terms  of  an  algebraic  expression  are  the  parts  con- 
nected by  the  signs   -f-  and  — . 

15.  An  algebraic  expression  is  called  a  Monomial,  a  Binomial, 
a  Trinomial,  a  Quadrinomial,  or  a  Polynomial,  according  as  it 
consists  of  one,  two,  three,  four,  or  many  terms. 

The  term  Polynomial  is  applied  also,  in  a  general  way,  to  any 
algebraic  expression  consisting  of  more  than  one  term. 

16.  Similar  Terms  are  terms  having  the  same  literal  quantities 
affected  with  the  same  exponents. 

Similar  terms  can  differ,  therefore,  only  in  their  numerical 
coefficients. 

17.  Positive  and  Negative  Quantities.  The  signs  -f  and  — , 
besides  indicating  the  operations  of  addition  and  subtraction,  are 
used  to  distinguish  quantities  of  opposite  character. 

Thus,  if  distance  east  is  considered  + ,  or  positive,  distance  west  must  be 
considered  — ,  or  negative.  If  gains  are  +,  losses  are  — .  If  temperature 
above  0  is  + ,  temperature  below  0  is  — .  If  the  latitude  of  a  place  north  of 
the  equator  is  -f ,  that  of  a  place  south  of  the  equator  is  — ,  while  that  of  a 
place  on  the  equator  is  0.  If  a  force  tending  to  move  a  body  in  one  direction 
is  +,  a  force  tending  to  move  it  in  the  opposite  direction  is  — .  And,  in 
general,  quantities  which  contribute  to  one  result  being  considered  positive, 
those  which  contribute  to  the  opposite  result  must  be  considered  negative. 

As  quantities  on  one  side  of  0  are  -f,  and  those  on  the  opposite 
side  of  0  are  — ,  positive  quantities  are  sometimes  said  to  be 
greater  than  0,  and  negative  quantities  less  than  0. 

For  example,  if  a  man  possesses  nothing  and  owes  one  hundred  dollars,  he 
must  earn  one  hundred  dollars  before  his  capital  can  be  expressed  by  0,  and 
we  say  he  has  a  hundred  dollars  less  than  nothing. 

An  increase  in  the  numerical  value  of  a  positive  quantity 
increases  its  algebraic  value;  but  an  increase  in  the  numerical 
value  of  a  negative  quantity  decreases  its  algebraic  value. 

18.  An  Axiom  is  a  truth  which  is  assumed  as  self-evident. 

19.  A  Theorem  is  a  formal  statement  of  a  truth  requiring 
proof. 


DEFINITIONS  AND  NOTATION  21 

20.  The  Hypothesis  of  a  theorem  consists  of  the  conditions  on 
which  it  is  affirmed. 

21.  A  Problem  is  a  question  proposed  for  solution. 

22.  The  Solution  of  a  problem  is  the  process  of  obtaining  the 
result  sought. 

23.  A  Rule  directs  how  to  proceed  in  solving  problems  that 
belong  to  the  same  class. 

24.  A  Demonstration  is  the  course  of  reasoning  by  which  the 
truth  of  a  theorem,  or  the  correctness  of  a  solution  or  a  rule,  is 
established. 

25.  A  Corollary  is  an  inference  from  a  preceding  theorem,  dem- 
onstration, or  solution. 

26.  A  Scholium  is  a  remark  upon  some  feature  of  what  has 
preceded. 


CHAPTER  II 
ADDITION 

27.  The  Algebraic  Sum  of  several  quantities  is  the  excess  of  the 
positive  over  the  negative,  or  of  the  negative  over  the  positive 
quantities. 

Thus,  if  a  man's  assets  are  $  5000  and  his  liabilities  $  3000,  the  excess  of 
his  assets  over  his  liabilities  is  $  2000.  Now,  if  we  consider  assets  positive 
and  liabilities  negative,  the  algebraic  sum  of  his  assets,  +  $  5000,  and  his 
liabilities,  -$3000,  is  +$2000. 

Again,  if  a  force  of  50  pounds  is  exerted  to  move  a  body  in  one  direction, 
and  a  force  of  30  pounds  to  move  it  in  the  opposite  direction,  the  effective 
force,  i.e.  the  aggregate  or  the  algebraic  sum  of  the  two  forces,  is  20  pounds. 

In  the  same  way  the  algebraic  sum  of  +  8,  +3,  —  5,  +  2,  and  —  4  is  4, 
and  the  algebraic  sum  of  +7,  —  6,  —  9,  and  +1  is  —  7. 

28.  Algebraic  Addition  is  the  process  of  finding  the  algebraic 
sum  of  several  quantities. 

29.  Prob.     To  add  monomials. 

Rule.  1st.  If  the  quantities  are  similar  {Art.  16),  find  the 
algebraic  sum  of  the  coefficients  and  annex  the  common  literal  part. 

2d.  If  the  quantities  are  dissimilar,  tvrite  them  in  succession  with 
their  signs  unchanged. 

Dem.     1st.   Let  it  be  required  to  add 

5  x*y,  7  x2y,  —  4  x?y,  and  —  2  x*y. 

By  Art.  9,  1st,  5  x2y  is  5  times  x*y,  and  7  x2y  is  7  times  the 
same  quantity.  Now,  5  times  any  quantity  and  7  times  the  same 
quantity  are  12  times  that  quantity,  giving,  in  this  case,  12  arfy. 
By  Art.  17,  —  ^x2y  is  so  opposed  in  character  to  +  12  arfy  as  to 
destroy  or  neutralize  4  of  the  12  times  v?y,  giving  -f  8  x2y.  Simi- 
larly, —  2  x2y  destroys  2  of  the  8  times  x2y,  giving  for  the  sum 
of  the  four  terms   +  6  x2y,  in  which  the  common  literal  part  is 

22 


ADDITION 


23 


simply  annexed  to  the  algebraic  sum  of  the  coefficients.      The 
same  reasoning  applies  to  any  other  set  of  quantities. 

2d.    Let  it  be  required  to  add  +3a,   +  4  6,  —  2  c,  and  —  d. 

Now,  3  a  and  4  b  are  not  respectively  3  and  4  times  the  same 
quantity ;  hence  the  sum  will  not  be  7  times  either  quantity,  and 
we  can  only  indicate  the  addition,  giving  for  the  sum  of  the  two 
3  a  +  4  b.  For  the  same  reason  the  addition  of  —  2  c  and  —  d  to 
this  sum  can  only  be  indicated,  the  minus  signs  being  preserved, 
since  these  negative  terms  neutralize  a  part  of  the  positive  sum. 
Hence  the  entire  sum  is 

3a  +  4:b-2c-d, 

in  which  the  given  quantities  are  written  in  succession  with  their 
signs  unchanged. 

30.  Cor.  Adding  a  negative  quantity  is  the  same  as  subtracting 
a  numerically  equal  positive  quantity. 

31.  Prob.      To  add  polynomials. 

Rule.  Write  the  quantities  so  that  similar  terms  shall  be  in  the 
same  vertical  column.  Add  the  columns  separately  and  connect 
their  sums  by  the  resulting  signs. 

Dem.  The  columns  may  be  added  separately  by  Art.  29,  1st. 
The  sums  being  dissimilar,  the  addition  of  them,  according  to 
Art.  29,  2d,  can  only  be  indicated  by  connecting  them  by  the 
resulting  signs. 

32.  In  writing  several  similar  terms  in  column  for  convenience 
in  adding,  much  time  is  saved  by  writing  the  literal  factors  only 
once.  The  plus  signs  also  may  be  omitted,  inasmuch  as  when  no 
sign  is  written,  the  plus  sign  is  understood. 

7  a%W 
-3 
-5 
2 
6 
9 


'  +  7  a3b2c* 

-  3  aWc* 

-  5  a3&2c* 

Instead  of    . 

+  2  «3fc2C* 

-  0  aVS-d 

+  9«3ft2c4 

+  4  a*b2c* 

'      7 

a36V     .     ' 

-3 

-5 

2 

or        { 

-6 

0 

[       4 

a86-2c* 

4  aWc* 


24  HIGHER  ALGEBRA 

While  the  first  operation,  exclusive  of  the  answer,  contains  48 
figures,  letters,  and  signs,  either  of  the  others  contains  but  15. 

33.  Note.  Although  the  principles  of  fractions  will  be  treated 
in  their  proper  place,  we  shall  from  the  beginning  assume  such 
knowledge  of  them  as  comes  from  the  study  of  Arithmetic. 

EXAMPLES  I 
Add  the  following : 

1.  2  a^  —  3  x2y  —  4  xy2  -f  6  y2,  4  x2y  —  xy2  —  5  a?3,  4  x2  —  5  x*y  —  y3 
+  2x>-3y2,  x2  +  8f-2y2  +  7xy2,  2  x*y  +  3  xy2  -  3  y*,  and 
2tf-4;ys  +  3y2  +  x2. 

Operation 

2  x2  -  3  x?y  -  4  xy1  +  6  y2 

4        -  1  -  5  x3 

4-5  -3  2-?/3 

1  7-2  8 

2  3  -3     • 

1  3  2-4 


8  x2  -  2  x2y  +  5  xy2  +  4  y°-  -     x* 

2.  m2  —  3  mn  -\-  2  n2,  3  n2  —  m2,  and  5  mn  —  3n2  +  2  m2. 

3.  2ab-3ax2-\-2a2x,    12  ab  +  10  ax2  -  6  a2x,    and    aar3-8a& 
—  5  a2#. 

4.  a?4  —  4  a^i/  -f  6  x?y2  —  4  sci/3  +  y4,    4  afy  —  12  x?y2  +  12  sc?/3  —  4  ?/4, 
6  #y  —  12  xy3  +  6  ?/4,  4  ajz/3  —  4  ?/4,  and  y\ 

5.  5  a2^  +  4  a2fo»2  +  m#y  and  10  a2  ex2  —  2  arbx2  4-  6  mx*y2. 

6.  3a62-4a26  +  a3,.5a&2-4ac2-c3,  and  2a26-  7a&2-  6ac. 

7.  2  a2  4-  5  ab  —  xy,     —  7  a2  +  3  ab  —  3  xy,     —  3  a2  —  1  ab  +  5  xy, 
and  9  a2  —  ab  —  2  xy. 

8.  5  aW  -  8  a2b3  +  x*y  +  ^2,  4  a263  -  7  a362  -  3  ^2  +  6  afy, 
3a3b2  +  3a2b3-3x2y  +  5xy2,    and    2  a263  -  a3b2  -  3  x*y  -  3  xy2. 

9.  a2-62+3a27;-5a&2,    3a2-4a2&+3  63-3a&2,    a3+b3+3a2b, 
2a3-4,b3-5ab2,  6a2b  +  10ab2,   and    -6a3-7 ci'b  +  Aab2 +  2b3. 

10.   2  m2  4-  ^  ?i2  —  |  m27i  4-  3  mn1  —  mn,  -|  n2  —  2  mn2  4-  3  m2?i, 

iii2  —  £  7i2  4-  4  mn  4-  J  m2w,  and  i  mna  4-  3  m2  —  3  mn  4-  2  w2. 


ADDITION  25 

11.  2a^-4x*  +  r5,  5  x?y  —  ab  +  x* ,  4ar-art,  and  2x$— 3+2  a£ 

12.  ax  +  2by  +  cz,  Vc+Vy  +  Vz,  _  3Vy  — 2VaJ  +  3Vs, 
4  cz  —  3  ax  —  2  by,    and    2  ax  —  4Vy  —  2Vz. 

34.  Literal  terms  that  are  similar  with  reference  to  only  part 
of  their  factors,  may  be  united  into  one  term  with  a  polynomial 
coefficient. 

Thus,  7  ax,  —  2  bx,  and  3  ex  are  similar  with  reference  to  x  only,  being, 
respectively,  7  a,  —  2  b,  and  3  c  times  x,  and  the  sum  is  (7  a  —  2  b  +  3  c) 
times  X,  or  (7  a  —  2  6  +  3  c)x. 

35.  Compound  terms  that  have  a  common  compound,  or  poly- 
nomial factor  may  be  added  with  reference  to  that  factor. 

Thus,  b(x2  —  y),  Z(x2  —  y),  and  —  2(x2  —  y)  are  similar  with  reference  to 
the  quantity  x2  —  y,  being,  respectively,  5,  3,  and  —  2  times  x2  —  y,  and  the 
sum  is  (5  +  3  —  2)  times  (x2  —  y),  or  6(x2  —  y). 

EXAMPLES  II 
Add  the  following: 

1.  ax,  bcx,  —  3bx,  —2  ex,  and  4  x. 

2.  ax  —  2 by,  2bx—3 by,  and  cy  —  ax. 

3.  ax  +  2  6y  —  4  z,  3  y  —  cz,  and  4  x  —  6  #  +  2  cz. 

4.  a  (x  -f  y)  -f  6  (x  —  y)  and  m  (x  +  y)  —  n  (x  —  y). 


5.   a  +  6 Vl  —  c2  and  a  —  b  Vl  —  c2. 


6.  5 V#  —  ?/,  5 a  Vc  —  y,—  3vx  —  y,  and  —  a  Ve  —  y. 

7.  | Va2  —  «,  —  -§ Va2  —  a?,  and  Va2  —  #. 

8.  ^L  +  ?_26?  i^-2  +  86,  and  --^-™_3&. 

Vie     2/  Vc     2/  Ve      2/ 


9.   (a+b-c)^/mi-x2,  (a-b  +  c)Vm2 -x2,  and  3aVm2-^. 


10.    (a  +  b  -  c)</xT^-f,  (a  -  b  +  c)W  -  ?,  and 

(b  +  c  —  c^Vic2  —  2/2. 


CHAPTER  III 
SUBTRACTION 

36.  The  Algebraic  Difference  of  two  quantities  is  their  difference 
with  regard  to  their  character  as  positive  and  negative,  the  order 
of  subtraction  being  independent  of  their  magnitudes. 

While  in  finding  the  arithmetical  difference  the  less  quantity  is 
always  taken  from  the  greater,  both  quantities  being  regarded  as 
positive,  in  finding  the  algebraic  difference  either  quantity  may 
be  taken  from  the  other,  and  one  or  both  of  the  quantities  may  be 
negative. 

Thus,  suppose  that  a  merchant  who  had  $  500  bought  goods  to  the  amount 
of  $  700,  and  we  are  asked  to  state  how  much  money  he  had  left.  In  an 
arithmetical  sense  the  question  is  absurd,  inasmuch  as  he  not  only  had 
nothing  left,  but  went  in  debt  $  200.  In  an  algebraic  sense,  however,  the 
question  is  not  absurd,  the  answer  being  —  $200. 

Again,  if  one  man  has  $  500  and  owes  nothing,  and  another  man  owes 
8200  and  has  nothing,  the  difference  of  their  possessions,  a  debt  being 
regarded  as  a  negative  possession,  is  $  700. 

37.  Algebraic  Subtraction  is  the  process  of  finding  the  algebraic 
difference  between  two  quantities. 

38.  It  is  seen  that  subtraction  is  the  inverse  of  addition,  the 
problem  being  to  find  one  of  two  quantities  when  the  other  and 
the  sum  of  the  two  are  given. 

39.  The  terms  Minuend  and  Subtrahend  are  used  in  Algebra  as 
in  Arithmetic,  but  Addition,  Subtraction,  Sum,  and  Difference  (or 
Remainder)  have  a  more  general  signification.  In  Arithmetic 
addition  always  produces  an  increase,  and  subtraction  a  decrease ; 
but  in  Algebra  addition  may  produce  a  decrease  and  subtraction 
an  increase. 

26 


SUBTRACTION  27 

40.  Prob.      To  subtract  one  quantity  from  another. 

Rule.  Conceive  the  signs  of  all  the  terms  of  the  subtrahend  to 
be  changed,  then  proceed  as  in  addition. 

Dem.     Let  it  be  required  to  subtract  m  —  n  from  a. 

If  the  whole  of  m  be  subtracted  from  a,  the  result  will  be 
a  —  m.  Now  since  the  quantity  subtracted  is  too  large  by  »,  the 
remainder  is  too  small  by  n :  hence  to  make  it  what  it  should  be, 
we  add  n  to  the  result,  giving  a  —  m  +  n.  In  this  operation  we 
see  that  we  have  changed  the  signs  of  the  subtrahend  and  added 
the  result  to  the  minuend. 

41.  Cor.  Subtracting  a  negative  quantity  is  the  same  as  adding 
a  numerically  equal  jyositive  quantity. 

EXAMPLES  III 
From  3  a3  -  4  a2b  +  5  ab2  -6  b3  take  a3  +  2  a2b  -  3  ab2  -  8  b\ 

Operation 
1  2        -3        -8 


2  a3  -  6  a26  +  8  ab2  +  2  b* 

2.  From  3x3-2x2-x-l  take  2x*-3a?  +  x  +  l. 

3.  From  x2  -f  2  xy  +  y2  take  x2  —  2  xy  -f  y2. 

4.  From  1  +  3x  +  3x2  +  x3  take  1  -  3z  +  3a2-  ic3. 

5.  From  10ar5-8a2  +  6a;  +  4  take  7  a-5  +  5ar>  -  3  x  +  7. 

6.  From  3  a  -  3  a2  +  1  +  a3  take  1  -  a3  -  a  -  a2. 

7.  From  a2  +  2  a&  -f  b2  take  the  sum  of   —  a2  -f-  2  at  —  62  and 
-2a2  +  2fc2. 

8.  From  2  x4  -|-  4  a3?/  —  5  x*y 2  —  3  .r?/3  -f- ?/4  take  x4  —  x?y  —  3  x?y2 
+  6xys  +  y\ 

9.  From  the  sum  of  2a—  3  6  +  4  d  and  2  6  +  4  c  —  3d  take 
the  sum  of  3c  —  4a  —  4  fr  —  2d  and  3 a  —  2 c. 

10.   From  1  +  3Vie  +  3a:  +  Va?  take  1  -  3 Vx  +  3  a;  —  Vx?. 


28  HIGHER  ALGEBRA 

11.  From  x^  -\-  2  x^if5  -f  yz  take  x5  —  2  x%y*  +  y* 

12.  From  5(x  +  y)  +  3(x-y)  take  2  0  +  y)  -  3  (a  —  y). 


'    13.    From  a  +  ?>Vl  —  c2  take  a  —  b  Vl  —  c2. 


14.  Subtract  l^a  +  ar'  from  J Va  +  a£ 

15.  Subtract  (b  —  a  —  c)  Vo2  +  ?/2  from  (a  —  &  +  (^Var2  4-  2/2. 


16.  Subtract  b  yjx  +  y  —  a^/x  —  y  from  dV#  +  y  +  &  V#  —  y. 

17.  What   must   be  added   to    4^—3^+2   to  produce   Ax3 
+  7x-6? 

18.  To  what  must   Xs  +  3  arfy  +  3<n/2  +  y3   be  added  to  produce 
3a?-x2y-2xy2  +  4:f? 

19.  What   must  be  subtracted  from    5a  —  46  +  3c    to  leave 
2a  +  2b-c? 


20.  From  what  must  V#  +  y  —  bvx  -\-y  —  d(x  —  y)  be  sub- 
tracted to  leave  5  V#  +  2/  +  a  V*  +  ?/  +  c  (x  —  if)  ? 

SIGNS   OF  AGGREGATION  AS  RELATED   TO   ADDITION 
AND   SUBTRACTION 

42.  Theorem.  1st.  A  parenthesis,  or  other  sign  of  aggregation, 
may  be  removed  without  changing  the  signs  of  the  inclosed  terms,  if 
it  is  preceded  by  +,  and  by  changing  the  signs  of  the  inclosed,  terms, 
if  it  is  preceded  by  — . 

2d.  Conversely,  terms  may  be  inclosed  within  a  parenthesis,  or 
other  sign  of  aggregation,  without  changing  their  signs,  if  it  is  pre- 
ceded by  ■{-,  and  by  changing  their  signs,  if  it  is  preceded  by  — . 

Dem.  1st.  A  sign  of  aggregation  indicates  that  the  inclosed 
terms  are  to  be  treated  as  a  whole  (Art.  11)  ;  hence,  if  preceded 
by  +  ,  all  the  inclosed  terms  are  to  be  added,  which  does  not 
change  their  signs  (Art.  29),  and  if  preceded  by  — ,  all  the  terms 
are  to  be  subtracted,  which  changes  their  signs  (Art.  40). 

2d.  The  truth  of  the  converse  is  seen  from  the  fact  that  the 
removal  of  the  sign  of  aggregation  by  the  preceding  principle 
would  restore  the  expression  to  its  original  form. 


SIGNS  OF  AGGREGATION  29 

43.  Sch.  When  signs  of  aggregation  occur  within  other  signs 
of  aggregation,  they  may  be  removed  in  succession,  and  it  is 
usually  expedient  to  begin  with  the  innermost. 


Thus,  a-  [b  +  {c-(d-e  -/)}] 

=  a-[6  +  {c-(d-e+/)}] 
=  a-  [6  +  {C-rt  +  e-/}] 
=  a-[b  +  c-d  +  e-f] 
=  a  —  b  —  c+d  —  e+f. 

EXAMPLES  IV 
Perform  the  following  indicated  operations : 

1.  2x2-\5x-(x2  +  2x-5)\. 

2.  4a-3&-{2&  +  3-(2a  +  &  +  l)}. 

3.  7o-[3o-  {4a-(5a-2)}]. 

4.  ^_z_(2a;-(3^  +  4-(z-l))). 


5.  2 m  —  [3 m  -[m-(2?»i-3m  +  4)|-(5m-  2)]. 

6.  4  z  -  (12x  -  2 y)  -  \2 z  -  (10  x  +  4y)- (2 x-6y)\. 

7.  x  -  [y  +  z  -  \x  -  (-  x  -  y)  +  z\]  +  \z  -  (2 x  -  y)\. 

In  the  following  unite  terms  with  reference  to  similar  factors, 
preceding  each  polynomial  coefficient  by  —  : 

8.  ax3  +  2bxi-x-3xi  +  4:x2  +  5x-cx2-2dx. 

9.  ax2  +  arx3  —  bx2  —  5  x2  —  co3. 

10.   oa3  —  2  c\c  —  [frc2  —  j  c»  —  c?a?  —  (dx3  +  3  ex2)  \  —  (ex2  —  bx)]. 


CHAPTER   IV 
MULTIPLICATION 

44.  Multiplication  is  the  process  of  finding  a  quantity  which 
shall  be  as  many  times  a  specified  quantity,  or  such  a  part  of  that 
quantity,  as  is  represented  by  a  specified  number. 

The  terms  Multiplicand,  Multiplier,  Product,  and  Factors  are 
used  as  in  Arithmetic. 

It  follows  from  the  definition  that  multiplication  by  a  positive 
number  is  the  repeated  addition  of  the  whole  or  a  part  of  the 
multiplicand  to  0 ;  while  multiplication  by  a  negative  number  is 
the  repeated  subtraction  of  the  whole  or  a  part  of  the  multiplicand 
from  0. 

771 

Thus,    a  x  b  is  a  repeated  b  times,  and    a  x  —  is  one  nth  part  of  a 

repeated  m  times  ;  while  a  x(—b)  is  a  subtracted  b  times,  and  a  x  (  —  —  ) 

is  one  nth  part  of  a  subtracted  m  times. 

The  only  difference  between    -  x  m    and  ax-    is  that  in  the  former 

n  n 

one  nth  part  of  a  is  represented  as  already  taken,  and  this  nth  part  is  to  be 

repeated  m  times  ;  while  in  the  latter  one  nth  part  of  a  is  first  to  be  taken, 

and  then  this  nth  part  is  to  be  repeated  m  times. 

45.  Theorem.  The  product  has  the  same  sign  as  the  multiplicand 
when  the  multiplier  is  positive,  and  the  opposite  sign  when  the 
multiplier  is  negative. 

Dem.  Repeated  addition  of  the  whole  or  a  part  of  the  multipli- 
cand would  not  change  the  sign  (Art.  29),  while  repeated  subtrac- 
tion of  the  whole  or  a  part  of  the  multiplicand  would  change  the 
sign  (Art.  40). 

46.  Cor.  i.  When  there  are  but  two  factors,  like  sig7is  give  +, 
and  unlike  signs  give  — . 

30 


MULTIPLICATION  31 

47.  Cor.  2.  The  product  of  an  even  number  of  negative  factors 
is  -f ,  and  of  an  odd  number  —. 

48.  Theorem.  The  exponent  of  a  quantity  in  the  product  is  the 
sum  of  the  exponents  of  the  same  quantity  in  the  multiplicand  and 
multiplier. 

Dem.  '  1st.   When  the  exponents  are  positive  integers. 
Let  it  be  required  to  multiply  am  by  a'\     By  definition  (Art. 
9,  2d) 

am  x  o"  =  aaaa  •  •  •  to  m  factors  x  aaaa  •  •  •  to  n  factors 
=  aaaaaa  •  •  •  to  m  +  n  factors  =  am+n. 

2d.   When  the  exponents  are  positive  fractions. 

TO  p 

Let  it  be  required  to  multiply  a"  by  aq. 

m  mq 

Now  aH  —  anq\  for  while  the  denominator  of  the  exponent  in 
the  latter  indicates  that  a  is  resolved  into  q  times  as  many 
factors  as  in  the  former,  the  numerator  indicates  that  q  times  as 

p  np 

many  of  these  factors  are  taken.     In  like  manner  aq  =  anq. 
Therefore, 

m  p  mq  np  1  1 

a"  x  aq  =  0^X0,^  =  (aFq)mq  x  (aFq)np, 
which,  by  the  first  part  of  the  demonstration,  is 

1  mq+np  m    p 

(an~qjmq+»P  =  a    »?    =  a«+9. 

3d.   When  the  exponents  are  negative. 
Let  it  be  required  to  multiply  a~m  by  a~n. 
By  definition  (Art.  9,  2d) 

a~m  x  or*  =  —  X  — 

am     an 

Now,  as  fractions  are  multiplied  by  multiplying  the  numera- 
tors together  for  a  new  numerator,  and  the  denominators  together 
for  a  new  denominator,  we  have  by  the  first  part  of  the  demon- 
stration, 

crw  x  a~n  =  —  x  —  =  -=-  =  a~n~n. 


32  HIGHER  ALGEBRA 

EXAMPLES  V 

Show  that  the  following  results  are  true,  whether  the  multipli- 
cation is  before  or  after  performing  the  operations  indicated  by 
the  exponents : 

l.  81*  x  81*  =  19683.  .  2.  16-*  x  16"*  =  fc 

3.   25~*  x  25^  =  1.  4.   a~2  x  a3  =  a. 

Find  the  products  of  the  following : 
5.  4  a2bs  and  5  a3b2.  6.   12  cr*bc2  and  6  a4b~2c. 

7.   14afy^  and  —3  x'hfK  8.    —  20a£y*  and  —  8aA/_i 

9.   5  a&2,   -  2  a2o3,  -  be,  and  3  a26c2. 

10.  3  ax,  —  3aV,  4  6?/,   —  ?/3,  and  2#y. 

11.  7  a?*   x*y,  —  2  a?2/3?  and  3  *%». 

12.  3a26-3c'",   -4a6c*,  2a~3oc-1,  and  9a362. 

13.  6afy*,    -5x2y,   —xnym,  3x~Y,  and  —  2x2my5n. 

49.  Prob.     To  multiply  one  polynomial  by  another. 

Kule.  Multiply  each  term  of  the  multiplicand  by  each  term  of 
the  multiplier  and  add  the  products. 

Dem.  Let  it  be  required  to  multiply  any  polynomial  by 
a-\-  b  —  c. 

If  we  multiply  it  by  a  and  then  by  b  and  add  the  results,  we 
shall  have  taken  it  a  +  b  times.  But  the  multiplier  required  that 
it  be  taken  only  a  +  b  minus  c  times.  Hence  from  a  +  b  times  the 
given  polynomial  we  must  deduct  c  times  the  same  polynomial. 
But  subtracting  this  product  is  the  same  as  adding  it  with  its 
signs  changed,  and  these  are  changed  by  regarding  the  minus 
sign  of  c  in  multiplying. 

50.  The  Degree  of  a  Term  is  the  sum  of  the  exponents  of  its 
literal  factors. 

51.  The  Degree  of  a  Polynomial  is  the  same  as  that  of  its  term 
of  highest  degree. 


MUL  TIPLICA  TION  33 

52.  A  polynomial  is  said  to  be  Arranged  with  reference  to  a  cer- 
tain letter  when  the  exponents  of  that  letter  increase  or  decrease 
in  the  successive  terms. 

53.  A  Homogeneous  Polynomial  is  one  in  which  all  the  terms 
are  of  the  same  degree. 

54.  Theorem.  The  product  of  two  homogeneous  polynomials  is 
itself  homogeneous. 

Dem.  If  the  degree  of  every  term  of  the  first  polynomial  is  m 
and  of  the  second  ?i,  then  the  product  of  any  term  of  the  first  by 
any  term  of  the  second  will  be  of  the  (m  +  n)th  degree.  Hence 
the  product  will  be  homogeneous. 

55.  If  multiplicand  and  multiplier  are  arranged  with  reference 
to  the  same  letter,  the  product  will  have  a  similar  arrangement, 
and  the  work  will  be  more  systematic  than  it  otherwise  would  be. 
When  terms  are  missing  from  an  arranged  multiplicand,  their 
places  should  be  supplied  by  terms  with  coefficient  zero.  For 
convenience  in  adding,  similar  terms  of  the  partial  products  are 
written  in  the  same  column.  The  literal  factors  should  be  ivritten 
but  once,  as  the  writing  of  a  coefficient  under  another  sufficiently 
indicates  that  the  literal  factors  are  the  same. 

EXAMPLES  VI 
Multiply  the  following : 

1.   6  x3  -  4  tfy  -  2  xy2  +  3  y3  by  2  x2  -  3  xy  +  4  y2. 

Operation 


6x3-    4x2y-    2xy2  +    3y3 
2x2-    Sxy  +    4  y2 

9xy* 

8 

12  x5-    $x*y-    4x3y2  +    6x2ys 
-  18             12               6          - 
24         -  16 

12  y5 

12  x5  -  20  x*y  +  32  xty2  -    4  x2y3  -  17  x\f  +  12  y* 

2.  3  a2  -  4  ab  +  6  b2  by  2  a2  -  5  ab  +  3  b2. 

3.  3  x2  +  2  x  -  4  by  x2  -  3  x  +  2. 

DOWNEY'S    ALG.  3 


34  HIGHER  ALGEBRA 

4.  2  ar3  +  4  -  3  x  -  3  x2  by  3  x  -  2  +  4  a* 

5.  #4  —  #y  4-  y4  by  ar*  4-  y2. 

6.  £2-M?/  +  2/2  by  tf  —  xy  +  y2. 

7.  m4  +  n4  -f  p4  —  m2n2  —  m2p2  —  n2p2  by  m2  +  ri2  4- 1>2. 

8.  am  —  an  +  a2  by  am  —  a. 

9.  4  a2  +  9  62  +  c2  +  3  6c  +  2  ac  -  6  a&  by  2  a  +  3  b  -  c. 

10.  a2  +  b2  4-  c2  —  ab  —  ac  —  6c  by  a  +  b  4-  c. 

11.  x4  -  2  x2  4-  3  x  -  5  by  x3  -  2  a2  +  3  x  -  4. 

Operation 

a440z3-2x243x  -    5 
a3_2z243;e  -4 


x1  4  0  x6  -  2  x5  +  3  a4  -    5  x3 

-2  4-6  10  x2 

3  _    6  9-15* 

-  4  8      -  12         20 

x7  -  2  x6  4     x5  +  3x4-  17a3  4  27  x'2  -  27  g  +  20 

12.  2  a4  4-  3  x2  -  4  by  x3  4-5  ar*  -  6  x  4-  3. 

13.  3  x4  -  4  afy2  -  2  ajy8  +  5 ?/4  by  2  x2  -  xy  4-  3  y2. 

14.  m5  —  7  m3  4~  4  m  —  8  by  m3  —  3  m  +  5. 

15.  *4  +  3*3-2*4-5by*3-2^4-7^-3. 

16.  a^  4-  4  ar3  -  3  by  3  x4  -  2  x2  +  4. 

17.  4  a*  —  3  x2f,y"  4-  6  a:p?/9  4-  2  t/3*  by  2  a^  4-  3  yq. 

MULTIPLICATION  BY   DETACHED  COEEFICIENTS 

56.  In  examples  like  most  of  the  foregoing,  in  which  the 
terms  of  both  multiplicand  and  multiplier  contain  the  same 
letters,  if  the  arrangement  be  made  the  same  in  both,  the  multi- 
plication can  be  effected  by  using  the  coefficients  alone,  since 
the  literal  factors  in  the  product  will  follow  the  same  law  of 
arrangement. 


MULTIPLICATION  BY  DETACHED   COEFFICIENTS       35 

EXAMPLES  Vn 
Multiply  the  following : 

1.  6  ar3  -  4  x*y  -  2  a#2  +  3  y3  by  2  x2  -  3  a#  +  4  #2. 

Operation 


6_    4-    2  + 

2-    3+    4 

3 

9 

8 

12-    8-    4 
-  18      12 

24- 

6 

6- 
16- 

12 

12-26      32-4-17      12 
Prod.,  12  xb  -  26  x*y  +  32  x3y2  -  4  xV  -  17  xy*  +  12  y5. 

2.   x*  +  2  a;  -  4  by  ar*  -  1. 

Operation 

1+0+2-4 
1+0-1 

10      2-4 

-1      0-2    4 

1      o      1-4-2    4 

Prod.,  x5  +  x8  -  4  x2  -  2  x  +  4. 

3.  3x2_a.4.2  by  3arJ+-2a;-2. 

4.  ar'-S^  +  Saj-l  by  af— 20  +  1 

5.  3a2  +  4:ax-5x2  by  2 a2 - 6 az  +  4 x*. 

6.  27arJ  +  9afy  +  3a*/2  +  ^  by  3a;-?/. 

7.  2ar}  +  6arJ-4a;-3  by  3^-4^  -0  +  5. 

8.  m3  +-  2  m2n  +  2  ra?i2  by  ?n,2  —  2  mn  +  2  n* 

9.  a;2  +  9a;  +  20,  0*-70  +  12,  and  x2-2x-15. 
10.  0*  — 3*4.1,  a^  +  aj  +  l,  and  af— «+l. 

When  the  student  writes  down  the  polynomials  to  be  multiplied, 
he  should  write  them  in  the  proper  position  for  multiplying,  and 


36  HIGHER   ALGEBRA 

then  not  make  the  unnecessary  repetition  of  detaching  the  coeffi- 
cients, but  proceed  at  once  as  follows  : 


6  x3  -    4x2y  -    2  xyt 

x2-    Sxy  +    4y2 

+ 

3  y* 

12-8        -    4 

-  18             12 

24 

6 

6 

16 

-  9 

-  8 

12 


12  x5  -  26  x4y  +  32  xsy2  -    4  x2y*  -  17  xy*  +  12  yb 

11.  ar5  -  2  ax2  +  2  a2z  -  3  a3  by  a2  -  3ax  +  2  a2. 

12.  #*+  6afy  +  12  xy2  +  8y3  by  a?-  3afy-f  3a#*  —  if. 

13.  aj4-3ar,  +  a2  +  4a;-6  by  2^-5^-3^  +  5. 

14.  or5-5a4  +  13ar3-a;2-a  +  2  by  «2-2»-2. 

SHORT  METHODS  OF   MULTIPLICATION 

57.  In  many  special  cases  the  process  of  multiplication  may  be 
made  very  short,  and  in  others  the  products  may  be  written  by 
inspection.  These  methods  are  of  so  much  importance  and  save 
so  much  labor  that  they  are  here  given  by  special  rules  and  theo- 
rems. The  student  should  make  himself  thoroughly  familiar  with 
them. 

58.  Prob.  To  multiply  by  a  binomial  of  the  first  degree,  the 
coefficient  of  whose  first  term  is  unity,  when  multiplicand  and  multi- 
plier contain  but  one  letter,  or  when  they  contain  two  letters  and  are 
homogeneous. 

Rule.  Arrange  the  multiplicand  in  descending  powers  of  the  first 
letter  of  the  multiplier,  supplying  the  places  of  any  missing  terms  by 
terms  with  coefficient  zero.  Call  the  numerical  part  of  the  second 
term  a;  multiply  the  coefficient  of  the  first  term  of  the  multiplicand 
by  a,  and  add  the  product  to  that  of  the  second;  multiply  the  coeffi- 
cient of  the  second  term  of  the  multiplicand  by  a,  and  add  the  prod- 
uct to  that  of  the  third,  and  so  on.  The  coefficient  of  the  first  term 
of  the  multiplicand  and  these  sums  in  order  will  be  the  coefficients 
of  the  product,  and  the  arrangement  of  the  letters  will  be  the  same 
as  in  the  multiplicand,  the  degree  being  one  greater. 


sin  HIT  METHODS  OF  MULTIPLICATION  37 

Dem.     Let  it  be  required  to  multiply 

3  x*  —  2  x"y  —  5  x 2y~  +  4  xif  —  6y4  by  x-\-3y. 

By  the  process  of  Ex.  1,  page  33,  we  have 

3  a;4  -  2  afy  -    5  afy2  +    4  ay5  -    6  yA 
x  +  3y 


3x?  —  2  x4y  —    5  x*y2  -f-    4  a*2?/3  —    6  #?/4 

9        _    6         -15  12        -18?/5 

3  ic5  +  7  afy  -  11  afy2  -  11  xFf  +    6  xy*  -  18  ^ 

Since  the  coefficient  of  the  first  term  of  the  multiplier  is,  by 
hypothesis,  unity,  the  coefficients  of  the  terms  of  the  first  partial 
product  will  always  be  the  same  as  those  of  the  multiplicand. 

The  coefficients  of  the  terms  of  the  second  partial  product  are 
a  (in  this  case  3)  times  the  coefficients  of  the  terms  of  the  multi- 
plicand, and,  being  placed  one  term  farther  to  the  right  to  bring 
similar  terms  together,  we  see  that  to  obtain  the  coefficients  of 
the  terms  of  the  product,  a  (3)  times  the  first  coefficient  of  the 
multiplicand  is  added  to  the  second,  a  (3)  times  the  second  is 
added  to  the  third,  and  so  on.  Omitting  the  repetition  of  the 
coefficients  of  the  terms  of  the  multiplicand  in  the  first  partial 
product,  the  work  stands  as  follows : 

(x  +  3y)(3x4-2x?y-    5a%2+    Axf-    G?/) 

9       _    c        -15  12        -18 

3         7       -11        -11  6        -18 

As  shown  before,  the  literal  factors  in  the  terms  of  the  product 
follow  the  same  law  of  arrangement  as  in  the  multiplicand,  and 
since  the  multiplier  is  of  the  first  degree,  the  degree  of  the  prod- 
uct is  one  greater  than  that  of  the  multiplicand.  Hence  supply- 
ing the  literal  parts  we  have  for  the  product, 

3  x5  +  7  x4y  -  1 1  xy  -llxtf  +  G^-lSy5. 

Only  the  a  of  the  multiplier  need  be  written ;  and  when  the  co- 
efficients are  small,  as  in  this  case,  or  so  related  that  the  sums  are 


38  HIGHER  ALGEBRA 

small,  the  additions  can  be  readily  made  without  writing  the  in- 
termediate products. 

(3)     3x4-2x3y-    5afy2  +    ±xtf-§y4 

7       -11        -11   ^        6      -18 

Thus,  3  times  3  are  9,  which  added  to  —  2  gives  7 ;  3  times 
—  2  are  —  6,  which  added  to  —  5  gives  —11;  3  times  —5  are 
— 15,  which  added  to  4  gives  — 11 ;  3  times  4  are  12,  which 
added  to  —  6  gives  6 ;  3  times  —  6  are  — 18,  which  added  to  0 
gives  — 18. 

Supplying  the  literal  parts,  we  have  for  the  product, 

3  x5  +  7  x4y  - 11  x*y2  -  11  afy8  +  6  xy4  - 18  y5. 

EXAMPLES  VIII 
Multiply  the  following : 

1.  4:X5-3x4y-Sxzy2  +  2x2f-6xy4-15y5  by  x-6y. 

Operation 

(-6)    4a;5-    3x*y-    8x*y2+   2x2ys-   6xy*  -15?/5 

-  27  10  50  -  18  21  90 

Prod.,  4  x*  -  27  xby  +  10  x4y2  +  50  x*y*  -  18  x2*/*  +  21  xy5  +  90  y6 

If  the  student  perform  the  multiplication  in  the  usual  way, 
writing  the  partial  products  in  full,  he  will  find  that,  between 
factors  and  product,  he  uses  70  figures,  letters,  and  signs,  while 
by  this  process  he  uses  but  14. 

2.  x3  +  4  x-y  +  7  xy2  —  5  y*  by  x  —  5y. 

3.  2 x* -  x*y  +  7 tftf -20 xf  +  75y4  by  x  +  4#. 

4.  4  m5  —  8  m4w  -f- 16  m3w2  — 16  m2nz  -f  8  mn4  —  4  »5  by  m  -j-  5  n. 

5.  3  m4  —  7  m2n2  +  6n4  by  m-\-6n. 

Operation 

(6)     3  m4  +    0  m*n  -  7  m%*  +    0  mnz  +6n4 

18-7  -42  6  36 


Prod.,  3  wi5  -(-  18  m*n  -  7  msn2  -  42  w%3  +  6  mn*  +  36  w1 


SHORT  METHODS  OF  MULTIPLICATION  39 

6.  5a6-6a42/2-4ay  +  7?/6  by  x-3y. 

7.  a4  —  4  x?y  +  3  xy3  —  9  ?/4  by  a  +  7#. 

8.  a4-4a3&  +  5a262  +  7afc3-12&4  by  a  +  12&. 

9.  3ar5-404  +  6a2-5a  +  lO  by  a  +  8. 

Operation 

(8)     3  x5  -    4  x4  +    0  x8  +  6  x2  -    5  x  +  10 

20-32  6  43      -30         80 

Prod.,  3  x6  +  20  x5  -  32  x*  +  6  x3  +  43  x2  -  30  x  +  80 

10.  2 a4  +  3^  +  40^  +  5 x  +  6  by  a +  5. 

11.  4a4-3a3-2a2-a-l  by  a-4. 

12.  ar5-3a4  +  5a8  +  2a2-6a;-8  by  a; +  7. 

13.  ^-8x2  +  5  by  «  +  6. 

14.  3  a;4 +  7  by  a;  — 9. 

15.  5  a5  —  5  by  a +  12. 

16.  ^  +  3^  +  6  by  a-10. 

59.  Successive  Multiplication.  When  several  factors  like  those 
of  Art.  58  are  to  be  combined  into  one  product,  the  literal  quanti- 
ties should  not  be  written  until  after  all  the  multiplications  have 
been  made. 

EXAMPLES  IX 
Find  the  products  of  the  following : 

1.  3a2-2a  +  l,  x  +  2,  x-1,  and  a  +  3. 

Operation 

(2)  3x2-    2x  +1 
(-1)  4-3  2 

(3)  1-7  5-2 

10      -  4      -  16  13-6 

Prod.,  3x5  +  10  x*-  4x*-16  x2  +  13  x-  6 

2.  a3-2a2  +  4a-8,  a  +  3,  and  a-4. 

3.  x2  —  3a +  5,  a  +  1,  and  a  — 7. 

4.  a  +  4,  a  —  3,  x-{-5,  and  a  —  2. 


40  HIGHER  ALGEBRA 

5.  2x^-3x2  +  Ax-7,  x  +  2,  x-3,  and  x  -f  4. 

6.  x  —  7,  x  +  1,  #  —  2,  ic  -f-  3,  a?  —  4. 

7.  3x2-Ax  +  2,  x  +  3,  x-5,  x-1,  x  +  2. 

8.  3  x  +  2>  3  ic  —  2,  x  +  4,  ic  —  5,  £  —  2,  and  x  +  1. 

9.  or2 +  2^  +  1,  x  +  3,  x  +  2,  x  —  2,  x  —  2,  and  x  —  1. 

60.  Theorem.  T%e  product  of  two  binomials  having  their  first 
terms  the  same  is  the  square  of  the  first  (or  common)  term,  plus  the 
algebraic  sum  of  the  second  terms  multiplied  by  the  first  (or  common) 
term,  plus  the  product  of  the  second  terms. 

Dem.  The  truth  of  the  proposition  is  seen  in  the  following 
results,  obtained  by  performing  in  the  usual  way  the  operations 
indicated : 

(a;  +  a)  (x  +  b)  =  x2  +  (a  +  b)  x  +  ab, 

(x  +  a)  (x  —  b)  =  x2  +  (a  —  b)  x  +  a  (—  b), 

(x  —  a)  (x  +  b)  =  x2  +  (b  —  a)  x  +  b (—  a), 

(x  —  a)  (x  —  b)  =  x2  +  (—  a  —  b)  x  +  (—  a)  (—  b). 

EXAMPLES  X 
Multiply  the  following : 

1.  x  +  8  by  x  —  5. 

Sug.  Omitting  the  intermediate  steps,  and  writing  at  once  the  result  by 
the  proposition,  we  have 

(x  +  S)(x  -  5)  =  x2  +  3 x  -  40. 

2.  x -  7  by  x  +  3.  Ans.  x2-±x-  21. 

3.  oj-15  by  #-4.  Am.  a;2 -19  #  +  60. 

4.  a  +  23  by  a  +  4.  Ans.  a2  +  27a  +  92. 

5.  x  +  10  by  x  -  6.  6.   x  -  12  by  x  +  8. 
7.  m  +  20  by  m  —  4.  8.   m  +  16  by  m  +  3. 
9.  x  —  11  by  x  -  7.  10.   y  -  22  by  ?/  +  4. 

11.  y  +  90  by  y  -  10.  12.   a  +  50  by  x  -  25. 


SHORT  METHODS   OF  MULTIPLICATION  41 

13.  z2  +  9  by  z2  -  6. 

Operation.     Here  z2  takes  the  place  of  x  in  the  demonstration,  and 

we  have 

(*2  +  9)  O2  -  6)  =  2*  +  3  z2  -  54. 

14.  a2  -  8  by  aj*  -  6.  15.   ^  -  12  by  ar*  +  7. 
16.   a?  +  14  by  a?  -  5.                   17.   z4  -  30  by  z4  +  5. 

18.  cub3  —  11  by  ax-3  +  5. 

Operation.      Here  ax3  takes'  the  place  of  x  in  the  demonstration,  and 
we  have 

(ax3  -  11)  (ax3  +  5)  =  a2x6  -6  ax9-  55. 

19.  «V  +  6  by  aV  -  8.  Ans.  a4x*  -  2  a?x2  -  48. 

20.  ?>iar>  -  13  by  mxA  +  3.  21.   wV  +  15  by  m2z4  -  6. 
22.  a¥  -  20  by  a¥  +■  4.            23.   a¥  +  23  by  a¥  -  3. 

24.  3^  +  6  by  3.^-4. 

Operation.     Here  3x2  takes  the  place  of  x  in  the  demonstration,  and 
we  have 

(3x2  +  6)(3  x2  -  4)  =  9x4  +  2(3  x2) -  24  =  9 x4  +  6 x2  -  24. 

25.  2*-9  by  2a  +  7.  26.  4^  +  11  by  4#2-8. 
27.   3z3-12  by  3z3  +  6.  28.  5z2-15  by  5z2-5. 

29.  x  +  7  y  by  x  —  3  ?/. 

Operation.     Here  7  y  and  —  3  y  take  the  places  of  a  and  6,  respectively, 
in  the  demonstration,  and  we  have 

(x  +  7  y)  (x  -  3  y)  =  x2  +  4  xy  -  21  y2. 

30.  ic  —  10  y  by  a  —  6  #.  31.  #  -f  12  y  by  x  +  4  #. 
32.  a;  —  9  y  by  a;  —  7  ?/.  33.  a^  +  8  y  by  ar*  —  4  #. 
34.  2z2-5?/  by  2z2  +  9y.  35.  z2  +  4/  by  z2-8y3. 
36.   z*-lly2  by  z3  +  7?/2.  37.  3a?  +  7tf  by  3ar3-22/2. 

38.  a;2  +  5  m  by  a^-2w. 

Operation.     Here  x2  takes  the  place  of  x  in  the  demonstration,  and  5  to 
and  —  2  n  of  a  and  &,  respectively,  and  we  have 

(x2  +  5  to)  (x2  -  2  n)  =  x4  +  (5  to  -  2  n)x2  -  10  win. 

39.  x  —  Sy  by  a;  +  3  z.  40.   a;2  -f  11  y  by  a^2  —  4  z. 

41.   Sx>+-ly  by  3ar>-5z.  42.  5a2  +  6m3  by  5a2-3n2. 


42  HIGHER  ALGEBRA 

61.  Theorem.  The  square  of  the  sum  of  two  quantities  is  equal 
to  the  square  of  the  first,  plus  twice  the  product  of  the  two,  plus  the 
square  of  the  second. 

Dem.     By  actual  multiplication,  we  have 

(x  +  y)2  =  (x  +  y)  0  +  y)  =  x2  +  2  xy  +  y2. 

62.  Theorem.  The  square  of  the  difference  of  two  quantities  is 
equal  to  the  square  of  the  first,  minus  twice  the  product  of  the  two, 
plus  the  square  of  the  second. 

Dem.     By  actual  multiplication,  we  have 

(x  -  y)2  =  (x  -  y)  (x  -  y)  =  x2  -  2  xy  +  y2. 

63.  Theorem.  TJie  product  of  the  sum  and  difference  of  two 
quantities  is  equal  to  the  difference  of  their  squares. 

Dem.     By  actual  multiplication,  we  have 

(as  +  y)  (x  -  y)  =  x2  -  y2. 


EXAMPLES  XI 

Square  the  following : 

1.   1+x. 

2. 

x  +  2. 

3. 

x2  +  y. 

4.   x  —  5  y. 

5. 

3  a  +  4  b. 

6. 

xn  +  2. 

m    x  ,  y 
7.    -  ±  £. 

y      x 

8. 

J£*-f  af*. 

9. 

xi  +  iy-t. 

10.   x2  +  y2z*. 

11. 

7x?-3y2. 

12. 

a*  —  |  a*b3. 

13.   x**  —  tf*. 

14. 

ax*  —  by2. 

15. 

5  a3xA  -j-  4  b2y3. 

16.   ±x  +  7xn+\ 

17. 

7  as*  -3y~\ 

18. 

2  am     3  bn  m 
3bn      2  am' 

Write  the  products  of  the  following : 

19.   x2  +  2  y  by  x2  —  2  y.  20.   3m2  +  5n3  by  3  m2—  5  n3. 

21.   1+f  a  by  1-f  a.  22.   5x3  +  3y2z  by  5x*-3fz. 

23.   2  #*  +  3  ^  by  2^  —  3  yK       24.   9  ax+  Vox  by  9  ax— aW. 


SHOUT  METHODS   OF  MULTIPLICATION  43 

64.  Theorem.     The  square  of  any  polynomial  is  the  sum  of  the 

squares  of  each  of  the  terms,  together  with  the  algebraic  sum  of  twice 
each  term  into  each  of  the  terms  that  follow  it. 

Dem.  When  there  are  more  than  two  terms,  part  of  them  may 
be  treated  as  constituting  the  first  term  and  the  others  as  the 
second  term  of  a  binomial.     Thus  : 

(a  +  b  +  c)»  =  [(a  +  b)  +  c]*  «  (a  +  h}*  +  2  (a  +  b)  c  +  c2 

=  a*  +  2ab  +  b2  +  2ac  +  2bc  +  c2, 

or  a2  +  b'2  +  c2  +  2  ab  +  2ac  +  2  be. 

Again, 

(a+b+c-d)2=[(a  +  b)i-(c-d)]2=(a  +  b)2+2(a+b)(c-d)  +  (c-(l)2 

=  a2+2ab  +  b2+2ac-2ad-2bd  +  c2-2cd+d2, 

or  a2+b2+c2+d2+2 ab  +  2  ac-2 ad  +  2  be  -2  bd-2  cd. 

By  inspecting  the  results  of  these  operations  we  deduce  the 
important,  time-saving  theorem  stated  above. 

EXAMPLES  XII 
Square  the  following : 

1.    a-b-  c.  2.  a2  —  b2  +  c. 

3.   m  —  n —  p  -f-  q.  4.  x2  -\-2  y  —  u  — •  3  v. 

5.   x*  +  a,*2  -f  a;  -f  1.  6.  a  —  b  +  c  —  cZ  -|-  e. 

7.   a;2  —  xy  +  y2.  8.  ar*  +  a^y  -f  a:?/2  -f-  y3. 

9.   3a^  — 5y  +  4**— 2*.  10.  a4  -  2  63  -  3  c2  +  4  tf. 

65.  Theorem.  The  product  of  the  sum  of  two  groups  of  terms 
by  the  difference  of  the  same  groups  is  equal  to  the  difference  of  the 
squares  of  the  groups. 

This  is  but  an  application  of  Art.  63,  regarding  the  first  group 
as  one  term  and  the  second  group  as  the  other  term  of  a 
binomial. 


44  HIGHER  ALGEBRA 

EXAMPLES  XIII 
Write  the  products  of  the  following : 
1.   a  +  b  —  c  by  a  —  b  +  c. 
Operation.     This  may  be  written, 
{a  +  (b  -  c)}{a  -  (b  -  c)}  =  a2-  (62  -  2  be  +  c2)  =  a2  -  ft2  +  2  6c  -  c2. 

2.  #  —  3  ?/  4-  5  z  by  x  —  3y  —  5  z. 

3.  a2  4-  y2  +  33/  by  x2  -\-y2  —  xy. 

4.  2x  —  ky  —  z  by  2  x  +  4  ?/  —  z. 

5.  4  a4  -  afy2  +  3  y*  by  4  a4  +  x*y2  4-  3  ?/4. 

6.  a*  —  x '4. 9  by  x2  -  x  -  9. 

7.  x2  +  x  -7  by  x2-x  +  7. 

8.  3  ar2  -  4  a;  4-  5  by  3  x2  +  4  a?  4-  5. 

9.  a  +  2&4-3c4-dbya  +  26-3c-riL 


CHAPTER  V 
DIVISION 

66.  Division  is  the  inverse  of  multiplication,  and  is,  therefore, 
the  process  of  finding  either  factor  when  the  other  factor  and  the 
product  are  given. 

The  terms  Dividend,  Divisor,  Quotient,  Remainder,  Numerator, 
and  Denominator  are  used  as  in  Arithmetic. 

67.  Theorem.  When  dividend  and  divisor  have  like  signs,  the 
quotient  is  +,  and  when  unlike,  —  . 

Dem.  If  d  is  the  divisor  and  q  the  quotient,  qd  is  the  divi- 
dend (Art.  66).  Now  by  the  law  of  signs  in  multiplication  (Art. 
46),  if  d  and  qd  have  like  signs,  q  is  +  ;  and  if  d  and  qd  have 
unlike  signs,  q  is  — . 

68.  Theorem.  The  exponent  of  a  quantity  in  the  quotient  is  the 
exponent  of  this  quantity  in  the  dividend,  diminished  by  its  exponent 
in  the  divisor. 

Dem.  Since  the  exponent  of  a  quantity  in  the  dividend, 
which  is  the  product  of  divisor  and  quotient,  is  the  sum  of  the 
exponents  of  that  quantity  in  these  factors  (Art.  48),  then,  con- 
versely, the  exponent  of  a  quantity  in  the  quotient  (one  of  the 
factors)  is  the  exponent  of  this  quantity  in  the  dividend  (the 
product),  diminished  by  its  exponent  in  the  divisor  (the  other 
factor). 

69.  Cor.  i.  Negative  exponents  arise  from  division  ivhen  the 
exponent  of  a  quantity  in  the  divisor  is  greater  than  that  of  the 
same  quantity  in  the  dividend. 

70.  Cor.  2.     Any  quantity  icith  the  exponent  0  is  1. 

Dem.  Let  x  represent  any  quantity  and  m  any  exponent. 
Now  xm  -~  xm  =  1.     But  by  the  law  of  exponents  just  established, 

xm  ~-xm  =  xm~m  =  x°.     Hence  x°  =  l. 
45 


46  HIGHER   ALGEBRA 

71.  Cor.  3.  A  factor  may  be  transferred  from  dividend  to 
divisor  (or  from  numerator  to  denominator  of  a  fraction,  ivhich  is 
the  same  thing),  and  vice  versa,  by  changing  the  sign  of  its  exponent. 

-n,  a        ax°      ax°~m     ax~m 

For 


bxm     bxm         b  b 

EXAMPLES  XIV 
Divide  the  following : 

1.   4  a5  by  2  a3.  2.   20  ax3  by  4  ax. 

3.   32  afyV  by  8  x?y4z5.  4.   27  xm  by  9  xn. 

5.   42  xm  by  7  x~n.  6.   a3^  by  aaA 

7.  a|  by  a2z.  8.   (ab)2m  by  (06)—. 

9.  18  or3  by  3x*.  10.  ra4af*  by  m2x~2. 

11.  a3  (a  ~  x)4  by  a3  (a  —  x)2. 

12.  12m"2(a2-3a;)3  by  3  m"3  (a2  -  3x)~2. 

13.  37  (or2  -  y2f  by  34  (oj2  -  y2)\ 

14.  a3(2a  +  4ar?)2n-3  by  a(2x  +  4ar5)"-3. 

Free  the  following  from  negative  exponents : 

a2b~3  _    3  afr"^-1^  4xmy~nz~p 

x-2f  '   5a~2d-3x-2  '      lx~nym 

72.  Theorem.  TJie  quotient  of  the  sum  or  difference  of  several 
quantities  is  equal  to  the  sum  or  difference  of  the  quotients,  the 
divisor  being  the  same. 

Dem.     Thus,   x  +  y  ~  z  is  equal  to  -  + 1  - *,  for  the  product 
n  n      71      n 

of  each  by  n  is  the  same,  viz.  x  -f  y  —  z. 

EXAMPLES  XV 
Divide  the  following : 

1.  6  a2b3  +  15  aAb2  -  12  a2b2  by  3  ab. 

2.  28  x?y4  -  84  x?y5  +  63  xY  by  7  ofy3. 


DIVISION  47 

3.  15  ax*  —  20  atx2  -+-  5  <&z  by   —  5  ax. 

4.  40  aH,c  -  24  aWc  -  32  abc*  by  8  abc. 

5.  9  x2"1  +  6  ar5'"  —  12  a4m  by  3  xm. 

6.  4  a"-*^-»  -  6  am+n68+n  by  2  a~nb\ 

7.  84  a1(V  -  12  a"1068  +  156  a2b6  by  12  a"10. 

8.  a*  -|-  3  a V  —  2  x$  by  ai 

10.   6  (x  -  2/)"-2  -  9  (x  -  y)n~l  +  12  (a  -  2/)M  by  3  (x  -  y)n~\ 

73.    Prob.     To  divide  one  polynomial  by  another. 

Rule.  Arrange  dividend  and  divisor  with  reference  to  the  same 
letter. 

For  the  first  term  of  the  quotient  divide  the  first  term  of  the  divi- 
dend by  the  first  term  of  the  divisor. 

Multiply  the  divisor  by  this  term  of  the  quotient,  and  subtract  the 
product  from  the  dividend. 

Treat  the  remainder  as  a  new  dividend  and  proceed  as  before, 
continuing  the  operation  until  there  is  no  remainder,  or  until  the 
first  term  of  the  remainder  is  not  divisible  by  the  first  term  of  the 
divisor. 

Dem.  Since  the  dividend  is  the  product  of  the  divisor  and 
quotient  (Art.  66),  that  term  of  the  dividend  which  has  the  high- 
est exponent  of  the  letter  of  arrangement  must  be  the  product  of 
those  terms  of  the  divisor  and  quotient  which  contain  the  highest 
exponents  of  the  same  letter.  Hence,  if  we  divide  the  first  term 
of  the  arranged  dividend  by  the  first  term  of  the  arranged  divi- 
sor, we  shall  obtain  the  first  term  of  the  quotient. 

If  the  product  of  the  whole  divisor  by  the  first  term  of  the 
quotient  be  subtracted  from  the  dividend,  the  remainder  must  be 
the  product  of  the  divisor  by  the  sum  of  all  the  other  terms  of 
the  quotient ;  hence  the  second  term  of  the  quotient  may  be  found 
from  this  new  dividend  as  the  first  was  found  from  the  original 
dividend. 

If,  finally,  there  is  no  remainder,  the  division  is  exact.  If,  in 
the  end,  there  is  a  remainder  whose  first  term  is  not  divisible  by 


48  HIGHER  ALGEBRA 

the  first  term  of  the  divisor,  the  quotient  may  be  completed,  if 
desired,  by  adding  this  remainder  over  the  divisor  in  the  form 
of  a  fraction. 

74.  Sch.  The  arrangement  of  the  terms  corresponds  with  the 
succession  of  thousands,  hundreds,  etc.,  in  decimal  numbers,  and 
these  operations  of  division  are  analogous  to  those  of  long  division 
in  Arithmetic. 

Example 

Divide  6  xA  -  13  ax*  +  13  a 2x?  -  13  a3.T  -  5  a4  by  2  x2  -  3  ax  -  a2. 


Operation 
6  x*  -  13  az3  +  13  a2x2  _  ^  asx  _  5  a4 
6  x*  -    9  ax*  -    3  a2x2 

\2x*- 
3a;2- 

-  3  ax  -  a? 

-  2  ax  +  5  a2 

-  4  ax3  +  16  a2x2  -  13  asx 

-  4axs+    6a¥+    2  a3x 

10  a2x2  -  15  aH  -  5  a4 
10  a2x2-  15a%-5a4 

75.  A  simple  inspection  will  show  that  the  following  may  be 
omitted  from  the  operation  as  entirely  unnecessary : 

1.  The  first  term  of  each  product  to  be  subtracted,  because  it 
is  always  the  same  as  the  first  term  of  that  from  which  it  is  to 
be  subtracted. 

2.  The  literal  factors  in  the  various  terms  of  the  products  and 
remainders,  because  they  are  always  the  same  as  those  of  the 
terms  of  the  dividend  under  which  they  stand. 

3.  The  bringing  down  of  terms  from  the  dividend  to  the  remain- 
ders, because  the  subtractions  can  just  as  well  be  made  while  they 
retain  their  original  positions. 

4.  The  signs  except  when  — ,  because  when  no  sign  is  written, 
the  +  sign  is  always  understood. 

With  these  omissions  the  above  operation  contracts  to  the  fol- 
lowing : 


6  x*  -  13  ax*  +  13  a2x2  -  13  a*x  -5  a* 

|  2  x'2  -  3  ax  -  a2 

-    9        -   3 

Sx2-2ax  +  5a? 

-    4             16 

6               2 

10          -15 

-15        -5 

SHORT  METHODS  OF  DIVISION  49 

When  terms  are  missing  from  the  dividend,  their  places  must 
be  supplied  by  terms  with  coefficient  zero. 

EXAMPLES  XVI 

Divide  the  following : 

1.    i  +  2x2-7x4-16x(i  by  1 +  2x  +  3arJ  +  4ar}. 

Operation 

-   I  +  Ox  +  2x2  +  Ox*  -  1  x*  +  Ox*  -  1G&  \  l  +  2x  +  3x2  +  4s8 


2 

3 

4 

-2 

-  1 

-4 

-4 

-6 

-8 

2 

1 

6 

9 

12 

-4 

-8 

-12 

-8 

-  12 

-16 

2.  15x4-32xi  +  50xi-32x  +  15  by  3^-4^  +  5. 

3.  6a*-31a?  +  23a?-2x-48  by  3^-5^  +  6. 

4.  2  a76  -  5  a662  -  11  a5b3  +  5  a464  -  26  a365  +  7  a2//  -  12  a&7  by 

SHORT  METHODS  OF   DIVISION 

76.  The  operation  of  division  may  be  still  further  shortened 
as  follows : 

If  the  signs  of  the  terms  of  the  divisor  after  the  first  are 
changed,  the  signs  of  those  terms  of  the  various  products  which 
were  subtracted  in  the  operations  above  will  be  changed,  and  the 
subtraction  changed  to  addition,  since  to  subtract  we  change  the 
signs  of  the  subtrahend  and  proceed  as  in  addition.  The  opera- 
tion at  the  bottom  of  page  48  would  then  become, 

%x^-13axi  +  13a2x2-13aix-baA\  2 x2  +  3 ax  +  a2 


9 
-4 

3 
16 

-   6 

-   2 

5 

3xi-2ax  +  5a2 

10 

-15 
15 

downey's  alg. — 4 


50  HIGHER   ALGEBRA 

It  will  be  seen  that  the  result  of  the  first  addition  to  the  term 
13  a2x2  is  not  used  in  obtaining  a  term  of  the  quotient,  and  that 
the  result  which  is  used,  viz.,  10  (with  a¥  understood),  can  just 
as  well  be  obtained  by  adding  13,  3,  and  —  6  without  first  adding 
13  and  3.  The  same  remarks  apply  to  the  additions  to  the  term 
—  13a3x.  With  these  omissions,  and  writing  the  terms  of  the 
products  as  near  as  may  be  to  the  corresponding  terms  of  the 
dividend,  the  operation  becomes, 

6  x4  - 13  ax3  +  13  a2x2  -  13  asx  -  5  a4\2x2  +  3ax  +  a2 
9  3-2  5      3a2-2az  +  5a2 

-4         -6  15 

10  0  0 

A  convenient  rule  for  this  operation  is  the  following : 

Arrange  dividend  and  divisor  with  reference  to  the  same  letter, 
supplying  the  places  of  any  missing  terms  of  the  dividend  by  terms 
with  zero  coefficients. 

Change  the  signs  of  all  the  terms  of  the  divisor  except  the  first. 

Divide  the  first  term  of  the  dividend  by  the  first  term  of  the  divisor 
for  the  first  term  of  the  quotient. 

Multiply  the  terms  of  the  changed  divisor  except  the  first  by  this 
term  of  the  quotient  and  write  the  product  under  the  corresponding 
terms  of  the  dividend,  omitting  the  literal  factors. 

Add  the  first  of  these  products  to  the  second  term  of  the  dividend, 
and  divide  this  sum  by  the  first  term  of  the  divisor  for  the  second 
term  of  the  quotient. 

Multiply  the  terms  of  the  changed  divisor  except  the  first  by  this 
second  term  of  the  quotient  and  write  the  products  under  the  corre- 
sponding terms  of  the  dividend,  omitting  the  literal  factors. 

Add  the  third  term  of  the  dividend  to  the  two  terms  that  stand 
under  it,  and  divide  this  sum  by  the  first  term  of  the  divisor  for  the 
third  term  of  the  quotient. 

Continue  this  operation  either  until  the  division  terminates  or 
until  one  of  these  sums  contains  a  lower  exponent  of  the  letter  of 
arrangement  than  the  first  term  of  the  divisor.  In  the  latter  case 
this  sum  and  the  sums  of  the  remaining  terms  of  the  dividend  and 
the  terms  that  stand  under  them  constitute  the  remainder. 


SHORT  METHODS  OF  DIVISION  51 

77.  This  is  a  modified  form  of  Horner's  Synthetic  Division  and 
should  be  thoroughly  mastered  by  the  student.  It  requires  but 
little  attention  and  practice  to  become  familiar  with  it,  while  the 
saving  in  time  is  very  great.  By  comparing  the  last  operation 
with  the  usual  process  as  given  in  the  first  solution  it  will  be 
seen  that  while  that  operation  (not  counting  dividend,  divisor, 
and  quotient)  contains  80  figures,  letters,  and  signs,  this  contains 
but  15. 

EXAMPLES  XVII 
Divide  the  following : 

1.  10 a4 -  27 «*a?  +  34  aV  -  18  ao8  -  8  a4  by  2a2-  Sax  +  ix2. 

2.  x4-Saxi-Sa2x2-\-lSatix-Sa4  by  x2  +  2ax-2a2. 

3.  4y5-24?/5  +  60t/4-80t/3  +  60?/2-242/  +  4  by  2?/2-4?/+2. 

4.  z6-5ar5  +  15a4-  24ar}  + 27z2- 13z  +  5  by  aj4 -  2 a8  +  4 aj2 
-2z  +  l. 

5.  2  a'b  -  5  a%2  -  11  a563  +  5  a4b4  -  26  o365  +  7  a266  -  12  a&7  by 
a4-4a3&  +  a262-3a&3. 

6.  a6-3aV  +  3«V-x6  by  a3 -3a2x  + Sax2 -x\ 

7.  9a?  +  24x2/  +  122/2  +  30a*  +  24yz  +  9z2  by  x  +  2y  +  3z. 

8.  8y5-22ay  +  20afy3  +  xy-7aty  +  6ar5  by  ±y2-3xy  +  2x2. 

9.  6ar5-7«42/  +  ary  +  20^-22^4  +  8?/5  by  2a2-3a>y+ ±y2. 

10.  l  +  2ar5  +  a;6  +  2a,'7  by  l  +  a;  +  x2. 

11.  16^  +  36^  +  81  by  4ar2-6x  +  9. 

12.  6a-20-2ar}  +  15arJ-a4+2arJ  by  5  +  2x*-±x-3x2. 

13.  12  a5  -  14  a46  -  10  a362  -  a263  -  8  a&4  +  4  &5  by  6a3-4a26 
-3a62  +  263. 

14.  14 xA  +  45 xhj  +  78  x2y2  +  45xf +  2ly4  by  2x>  +  5xy  +  ly2. 

15.  1  -  a6  by  1  +  2  a;  +  2  a*  +  arl 


Operation 

l+Os  +  O^  +  O^  +  Oa^  +  Oz5- 

£6|l-2x-2x2-x8 

-2     -2      -1          2      -2 

1      1  -  2  x  +  2  x2  -  x8 

-2         4          4-4          2 

2-4          2 

-10          0 

0 

52  HIGHER  ALGEBRA 

16.  xs  -|-  if  by  x4  —  xhj  -|-  x2?/2  —  &y3  -J-  if. 

17.  a8  +  a?b-  +  a464  +  a266  +  68  by  a4  +  a?b  +  a262  +  a&3  +  b4. 

18.  a6-21ary  +  24a;?/5-8y5  by  rf-Sxy  +  if. 

Sug.  From  the  terms  given,  the  arrangement  is  seen  to  be  with  reference 
to  the  descending  powers  of  one  letter  and  the  ascending  powers  of  the  other. 
The  dividend  should,  therefore,  be  written 

se6  +  0  xby  +  0  x*y2  -  21  xsy*  -f  0  x2y*  +  24  xyh  -  8  y*. 

19.  Aa4_jas_  Ja*  +  |a  +  v  by  |a2-a-f. 

20.  36^-6^-4^  +  i2/-hi2/2  +  J  by  6*~}y-'$ 

21.  a&0*  +  2  (a  -  6)  iB3  -  (a2  +  4  -  b2)  x2  +  2  (a  +  b)  x  -  ab  by 
bx*  +  2  x  —  a. 

22.  6  a4  -  25  afy  4- 16  x*y2  -17 xif +  3oy4  by  2 x-7 y. 

Sug.  When  the  divisor  is  a  binomial,  as  in  this  example,  the  operation 
takes  the  following  form : 

6  x4  -  25  x*y  +  16  s2?/2  -  17  xy*  +  35  y4  |  2  a;  +  7  y 


21        -  14  7        -  35        3  xs  -  2  x2y  +  zy2  -  5  ya 


-    4  2-10  0 

23.  a  4-  9  x*  -  1  4-  3  x2  by  3  a?  -  1. 

24.  7  a?  -  24  x2  4-  58  a  -  21  by  7  a  -  3. 

25.  2  a4  4-  27  aft8  -  81  64  by  a  +  3  6. 

26.  6  or5  -  11  or*  4-  3  x2  4-  4  x  -  4  by  3  a2  -  4. 

27.  6  a4  -  96  by  3  a  -  6. 

28.  a1+n  +  aM6  4-  abn  4-  &1+n  by  a*  4-  bn. 

29.  mm+1  4-  nmm  -\-  amnan  -f  cman+1  by  m  -f  w. 

30.  6  a4-  13  a36  4-  9  a2b2  +  10  a&3  -  8  b4  by  3  a  -  2  6. 

31.  10ar,-7ic42/  +  8«3?/2-9xy+10^4  +  28  2/5  by  5x  +  4y. 

78.    "R7ien  ^e  coefficient  of  the  first  term  of  a  binomial  divisor  of 
the  first  degree  is  unity,  the  operation  becomes  exceedingly  simple. 


SHORT  METHODS   OF  DIVISION  53 

The  process  is  of  so  much  importance  and  is  of  such  frequent 
occurrence  in  subsequent  parts  of  the  work  that  a  special  rule  is 
here  given. 

79.  Prob.  To  divide  by  a  binomial  of  the  first  degree  when  the 
coefficient  of  the  first  term  is  unity. 

Rule.  Call  the  coefficient  of  the  second  term  of  the  divisor,  with 
its  sign  changed,  a;  multiply  the  coefficient  of  the  first  term  of  the 
dividend  by  a,  and  add  the  product  to  that  of  the  second;  multiply 
this  sum  by  a,  and  add  the  product  to  the  coefficient  of  the  third  term 
of  the  dividend,  and  so  on.  When  the  last  sum  is  zero,  the  division 
is  exact;  otherwise,  the  last  sum,  with  the  literal  part  of  the  last 
term  of  the  dividend,  will  be  the  remainder,  and  the  coefficient  of  the 
first  term  of  the  dividend,  and  the  other  sums  in  order,  will  be  the 
coefficients  of  the  quotient,  the  arrangement  of  the  letters  of  the  quo- 
tient being  the  same  as  in  the  dividend,  and  the  degree  being  one  less 
than  that  of  the  dividend. 

Note.  This  rule  may  be  deduced  as  a  special  case  of  the  last 
process ;  but  it  is  deduced  below  directly  from  the  ordinary  proc- 
ess of  "  long  division,"  with  which  the  student  has  become  famil- 
iar in  Arithmetic  and  Elementary  Algebra,  in  order  that  the 
relation  to  that  process,  and  at  the  same  time  the  great  saving  in 
labor  and  time,  may  be  seen. 

Dem.     By  the  ordinary  process  of  "  long  division  "  we  have, 

2 3^-3 ^y-13a,Y+14aY-2 ^-12^1    x  -Sy 
2x5-6x4y  2xA+3xiy-4:X2y2 

3x<y-13x>y*  +2a*»+4jf 

Sx4y-  9x*y2 

-  Ixy+Ux2!? 

-  4a?if +12 aty 


2x>tf-2xyA 
2x?y3-6xy4 


±xy4-\2tf 
±xy4-\2tf 


54 


HIGHER  ALGEBRA 


Omitting  the  wholly  unnecessary  parts,  without  changing  the 
position  of  the  parts  that  remain,  this  becomes, 


2  a5  -3 

-6 

3 


x4y  -  13 

-9 
-4 


xY  +  14 


12 

2 


ry 


-6 
4 


a^4  - 12 


12 

0 


y5  |  a?-3y 
2x4  +  3x3y-±x2y2 
+  2xtf  +  ±y4 


Now  since  the  coefficient  of  the  first  term  of  the  divisor  is,  by 
hypothesis,  unity,  the  coefficients  of  the  terms  of  the  quotient  will 
be  the  same  as  the  coefficients  of  those  terms  which  are  divided  by 
x  to  obtain  them  (in  this  case  the  same  as  in  the  terms  2  x5,  3  x4y, 
—  4  arty2,  2#y,  and  4:xy4).  Hence  the  coefficient  of  the  first  term 
of  the  dividend  and  the  coefficients  of  the  various  remainders  in 
order  will  be  the  coefficients  of  the  terms  of  the  quotient.  Any 
one  of  these  remainders  is  found  by  multiplying  the  one  before 
it  (which  is  the  same  as  the  coefficient  of  the  corresponding  term 
of  the  quotient)  by  the  coefficient  of  the  second  term  of  the  divi- 
sor, and  subtracting  the  product  from  the  coefficient  of  the  next 
term  of  the  dividend.  By  changing  the  sign  of  the  second  term 
of  the  divisor  this  subtraction  will  be  changed  to  addition.  The 
degree  of  the  quotient  is  one  less  than  that  of  the  dividend, 
because  the  divisor  is  of  the  first  degree.  Hence  writing  in  place 
of  the  divisor  only  the  coefficient  of  the  second  term  with  its  sign 
changed,  omitting  from  its  usual  place  the  quotient,  since  its 
coefficients  are  but  a  repetition  of  numbers  written  before,  and 
writing  the  products  and  sums  as  near  as  may  be  to  the  terms  of 
the  dividend  with  which  they  belong,  the  operation  becomes, 


2x?-3x4y-  13  xY  +  14  tftf 
6  9-12 


3        -    4 


2  xy4  -  12  y5  [£ 
6  12 

4  0 


SHORT  METHODS   OF  DIVISION  55 

Supplying  now  the  literal  parts,  we  have  for  the  quotient, 

2  x4  +  3  afy  -  4  x2y2  +  2  xif  +  4  y4. 

When  the  coefficients  are  small,  as  in  this  case,  or  so  related 
that  the  sums  are  small,  the  additions  can  be  readily  made  with- 
out writing  the  products. 

2  x6  -  3  x4y  -  13  x%f  +  14  xhf  -  2  xyA  -  12  t/5  |_3 
3-4  2  40 

Thus,  3  times  2  are  6,  which  added  to  —  3  gives  3 ;  3  times  3 
are  9,  which  added  to  — 13  gives  —  4 ;  3  times  —  4  are  — 12, 
which  added  to  14  gives  2;  3  times  2  are  6,  which  added  to  —2 
gives  4 ;  3  times  4  are  12,  which  added  to  —  12  gives  0.  Supply- 
ing the  literal  parts,  we  have  for  the  quotient, 

2  x4  +  3  x%y  -4:X2y2  +  2xif  +  4:  y\ 

When  the  last  sum  is  anything  other  than  zero,  this  sum 
(including  the  literal  part)  is  the  remainder,  since  the  operation 
by  which  it  is  obtained  is  the  same  as  subtracting  the  product  of 
the  last  terms  of  the  divisor  and  quotient  from  the  last  term  of 
the  dividend. 

80.  By  comparing  the  last  operation  with  the  usual  process 
by  "  long  division,"  as  given  in  the  first  operation,  it  will  be  seen 
that  while  that  operation  (not  counting  dividend,  divisor,  and 
quotient)  contains  93  figures,  letters,  and  signs,  this  contains 
but  6. 

EXAMPLES  XVIII 

Divide  the  following : 

1.  x*  —  11  x?y  +  41  xy2  —  50  y8  by  x  —  5y. 

Operation 

x*  -  11  xhj  +  41  xy2  -  50  if  [5 
-    6  11  5 

1,  —  6,  and  11  are  the  coefficients  of  the  terms  of  the  quotient 
and  5  is  the  coefficient  of  the  remainder. 

Ans.  x2  —  6  xy  -f- 11  y2,  with  remainder  5  y\ 


56  HIGHER  ALGEBRA 

2.  7ar5-3afy-2a2/2-40.v3  by  x-2y. 

3.  x4  +  4:Xiy-13xiy2-28xy3  +  60y4  by  x-2y. 

4.  2x4  +  llxiy  +  9x2y2-8xy3  +  16y4  by  *  +  4y. 

5.  12  x4  -  48  arty  +  11  arfy2- 45  ay^  7  ?/4  by  #-4?/. 

6.  xi  +  4x4y-2xiy2-18x2y!i-6xy*  +  9y5  by  x  +  3y. 

7.  3ar,  +  2xV-21a%2-14ary}  +  36ay  +  24?/5  by  a-2y. 
a  3^-25^i/-15ar?2/2-31ic22/5  +  38^4-18^  by  x-9y. 
9.  4x6-24arV+23xy-14arY+afy4-25ay5-25?/6  by  x-5y. 

10.  x6-2x'V-45xy+22arY+24a,V+2ay+12?/6  by  x+6y. 

11.  2ar5-9x4  +  llar3-20x  +  6  by  x  -  3. 

Operation 

2  x5  -  9  x4  +  11  x3  +  0  x2  -  20x  +  6  |_3 

-3  2  6-20 

<4*a.  2  x4  -  3  x3  +  2  x2  +  6  x  -  2. 

12.  a;4  +  ar3  -  50  a2  -  40  x  - 14  by  *  -  7. 

13.  3^  +  85^-3^-45  by  x  +  3. 

14.  8^-8^-60^  +  400  by  a? +  4. 

15.  5aj4-46ar}  +  48ar2  +  7a;-56  by  a; -8. 

16.  a? -20 a?- 20a?2 -20 a; -25  by  a?-5. 

17.  a*-13a**  +  15a?-30a*-75a;  +  36  by  a; -12. 

18.  2x4  +  35ar5  +  80arJ  +  75a;  by  a; +  15.     - 

19.  a^  +  3ar5-7ar4-10ar3  +  6x2+9a;-10  by  x-2. 

20.  ^-25^  +  47  arJ-21ar2-43a;-69  by  a; -23. 

21.  ^-64  by  x-2. 

Operation 

x6  +  0  x6  +  0  x*  +  0  x3  +    0  x2  +    0  x  -  64  |_2 
2  4  8  16         32  0 

Ans.  x5  +  2  x4  +  4  x3  +  8  x2  +  16  x  +  32. 

22.  a;4— 81  by  x  —  3. 

23.  x^  +  if  by  x+?/. 

24.  a6  —  729 1/6  by  x—3y. 


SHORT  METHODS  OF  DIVISION  57 

81.  Successive  Division.  In  dividing  successively  by  several 
binomials  of  the  first  degree  when  the  coefficient  of  the  first  term 
of  each  is  unity,  the  letters  should  not  be  written  until  after  the 
last  division.  It  must  be  noted  that  the  degree  is  reduced  by 
one  for  each  division. 

EXAMPLES  XIX 
Divide  successively  the  following : 

1.  x5-3z4-9arJ  +  21ar2-10z  +  24  by  a._2j  x  +  3,  and  x  —  4 

Operation 

xb  _  3  Xi  _    9  X3  +  21  &  -  10  x  +  24  | 2 

_  1      _  ii      _    i      _  12  0  J  -3 

-4  1-4  0  | 4 

0  10  j^i  X2  +  o  x  +  1,  or  x-  +  1. 

2.  a4-ar5-19ar>  +  49#-30  by  x-3  and  z  +  5. 

3.  x*  +  5xi-30xi-80x  +  224:  by  »  — 2,  #-4,  and  x  +  4, 

4.  ars  +  z4-31arJ  +  68arJ-54a;  +  63  by  x-3  and  x  +  7. 

5.  ^-7^4-5 ^+55^-126^+72  by  *-l,  a?-2,  z-3,  z+3, 
and  a;  — 4. 

6.  x6-  14  a;4  +  49^-36  by  »— 1,  a+1,  z-2,  z+2,  and  x-3. 


Operation 

x6  +  0z5-14x4  + 

0«3 

+  49z2+    Ox- 

-36  [ 

i 

1       -13      - 

13 

36          36 

0  I 

-l 

0      -13 

0 

36            0 

! 

2 

2      -    9      - 

18 

0 

1 

-2 

0-9 

0 

3 

3            0 

Ans.  x  -f  3, 

7.  x5  - 10  x4  -  3  ar* +258  x2  -  550  «  -  200  successively  by  x  -  4, 
x  —  5,  and  x  +  5. 

82.  Theorem.  fTAen  the  dividend  is  the  square  of  the  first  term 
of  the  divisor,  plus  or  minus  twice  the  product  of  the  first  and  second 
terms,  plus  the  square  of  the  second,  the  quotient  is  the  same  as  the 
divisor. 

This  is  a  consequence  of  Arts.  61  and  62. 


58  HIGHER  ALGEBRA 

83.  Theorem.  When  the  divisor  is  the  sum  or  difference  of  two 
quantities  arid  the  dividend  is  the  difference  of  the  squares  of  these 
quantities,  the  quotient  is  the  difference  or  sum  of  the  same  quantities. 

This  is  a  consequence  of  Art.  63. 

EXAMPLES  XX 

Divide  the  following : 

1.  S6x4^-36xy  +  df  by  6^  +  3^. 

2.  49 a8 -  70 a4b5  +  25 b10  by  7 a4-  5b5. 

3.  16a6-64fc4  by  4a3-8&2. 

4.  Six5  — Ay2  by  9x%  +  2y. 

5.  121a6-264ary+144y2  by  11  Xs  -12/. 

6.  64  a4b2  +  112  a?b(?  +  49  a2c*  by  8  a2b  +  7  ac3. 

84.  TF/ien  the  divisor  is  readily  separated  into  binomial  factors  of 
the  first  degree  having  unity  for  the  coefficient  of  the  first  term  of  each, 
the  division  is  most  expeditiously  made  by  dividing  successively 
by  the  factors,  as  in  the  process  of  Art.  81. 

EXAMPLES  XXI 

Divide  the  following : 

1.  ^  + 3a;4- 20ar3-60«2+ 64^  +  192  by  ^-6^  +  8. 

Operations 

The  factors  of  x2  —  6  x  +  8  being  x  —  2  and  x  —  4,  we  divide  successively 
by  these  factors,  thus  : 

5c5  +  3  x4  -  20  x>  -  60  x2  +  64  x  +  192  [2 
5-10-80-96  0[4 

9  26  24  0 

.4ms.  x3  +  9  z2  +  26  x  +  24. 

2.  ^4-3x3-7^+15x  +  18  by  a2-6a?  +  9. 

3.  ^  +  3x3_7a.2_27^_i8  by  x2-9. 

4.  0?  + 3 as4 -17 a?-  27^2+52x  +  60  by  a?  +  2x-15. 

5.  o?6-6;c4-4ar?+9.T2+12a,'  +  4  by  ^-4^  +  4. 

6.  ^-7^  +  17^-28^+37^-20  by  x2-5x  +  4. 

7.  3x5  +  x4-5xi-6x2-17x-U  by  x2-x-2. 


CHAPTER   VI 
FACTORING 

85.  A  Factor  of  a  quantity  is  a  quantity  that  will  divide  it 
without  a  remainder. 

86.  The  Factors  of  a  quantity  are  those  quantities  which  multi- 
plied together  produce  it. 

87.  To  Factor  a  quantity  is  to  separate  it  into  its  factors. 

88.  A  Prime  Quantity  is  one  which  has  no  integral  factors 
except  itself  and  unity. 

89.  A  Composite  Quantity  is  one  which  has  integral  factors 
other  than  itself  and  unity. 

90.  Theorem.  Any  monomial  factor  which  occurs  in  every  term 
of  a  polynomial  can  be  removed  by  dividing  each  term  of  the  polyno- 
mial by  it. 

Thus,  a3x  +  a2x2  —  a¥  =  a2x  («  +  «-  cfix2). 

91.  Theorem.  If  two  terms  of  a  trinomial  are  p>ositive,  and  the 
remaining  term  is  ±  twice  the  product  of  their  square  roots,  the  tri- 
nomial is  the  square  of  the  sum  or  difference  of  these  square  roots. 

This  is  a  consequence  of  Arts.  61  and  62. 

92.  Theorem.  The  difference  between  two  quantities  is  equal  to 
the  product  of  the  sum  and  difference  of  their  square  roots. 

This  is  a  consequence  of  Art.  63. 

93.  Theorem.  If  a  polynomial  of  six  terms  consists  of  three 
perfect  squares  and  three  double  products  of  their  square  roots, 
taken  in  pairs,  the  polynomial  is  the  square  of  the  algebraic  sum  of 
these  sq)iare  roots. 

This  is  a  consequence  of  Art.  64. 

Arts.  169  and  170  furnish  the  best  means  of  factoring  poly- 
nomials that  are  perfect  squares. 

69 


60  HIGHER  ALGEBRA 

EXAMPLES  XXII 

Factor  the  following : 
1.   a4&  +  3a664  +  4a3&3.  2.   3a4-6a5-12a2  +  9a3. 

3.  12  a4b7  -  20  a3b*  + 16  ab7- 8  a5bs. 

4.  10  xHf  +  5  fcty*  —  15  a?%*  —  20  as*y*. 

5.   a2  +  6«&-f9  62.  6.   a4  -  $  a?b  +  16  b2. 

7.   a^  —  afy  +  Jy2.  8.   x4  —  Ay2. 

9.   3ar  —  48  2/4.  10.   Sx'-lSxy4. 

11.   a4-?/8.  12.   81  A2 -1. 

13.   1  —  10  xy2  -f  25  x*y4.  14.   4  m2#4  +  4  mnxPy  -f  n?i/2. 

15.    (^  +  2/)2-f2(^  +  2/)  +  l-  16.   x>-(x-y)\ 

17.   a2-fc2-c2  +  26c.  18.    (a?  +  2/)4  +  4(x  +  2/)2  +  4. 

19.  (a2  +  2/)2-6V(a2  +  2/)+9z2. 

20.  (2x2  +  3^-8)2-(^-3^-8)2. 

21.  a2  +  b2  +  c2-2ab  +  2ac-2bc. 

22.  ^  +  !/2  +  9  +  2^?/  — 6a;  — 62/. 

23.  4  x2  —  4  #1/  +  y2  -f- 12  #z  —  6  2/2  +  9  z2. 

24.  9a4  +  4?/2-f  16z2-  12x2y-2Ax2z  +  16yz. 

25.  16  x6  +  24  a«y  +  9  y4  -  40  ^  -  30  y2z  +  25  z2. 

26.  One  factor  of  x4 —  7  a? +  6x2 +  23x  —  15  being  #  -  5,  what 
is  the  other  ? 

Sug.     Divide  as  in  Art.  79. 

27.  One  factor  of  Xs  —  6  x2y  + 15  #?/2  —  18 1/3  being  x  —  Sy,  what 
is  the  other  ? 

28.  Two   factors   of    2x4 -\-5x?  —  13x2 +  16    being   a?  +  4   and 
a?  -h  1,  what  is  the  third  ? 

29.  One  factor  of  Xs  —  Ax2  —  35 #  + 150  being  a?  +  6,  what  are 
the  other  two  ? 

30.  One  factor  of  4  Xs  —  32  a^  +  69  x  —  45  being  a?  —  5,  what  are 
the  other  two  ? 


FACTORING  61 

94.    Prob.     To  factor  a  trinomial  of  the  form  a*2"*  +pxm  +  q. 

Solution.     We  have  seen  (Art.  60)  that 

(xm  +  a)  (af*  +  b)=  x2"1  +  (a  +  b)  of  +  ab. 

Hence  if  in  an  expression  of  the  form  x2™  +  pxm  +  q,  q  can  be 
resolved  into  two  factors,  as  a  and  b,  snch  that  their  sum,  a  +  &, 
is  equal  to  p,  then  the  factors  are  xm  +  a  and  xm  +  6. 

EXAMPLES  XXIII 
Factor  the  following : 

1.  a2 +  7  a; +  10. 

Here  7  a;  is  the  product  of  the  square  root  of  the  first  term,  and  the  sum 
of  two  factors,  2  and  5,  of  the  last  term.  Hence  the  factors  of  x2  +  Ix  -f  10 
are  x  +  2  and  x  +  5. 

2.  ^  +  4^-21. 

Here  4  a;2  is  the  product  of  the  square  root  of  the  first  term,  and  the  sum 
of  two  factors,  7  and  —  3,  of  the  last  term.  Hence  the  factors  of  oj4+4x2— 21 
are  x  +  7  and  x  —  3. 

3.  ^_4aj_32.  4.  #2-8a  +  15. 
5.  ^  +  10^  +  9.  6.  x2  +  6x-72. 
7.  a* +  12  a; -45.  8.  ar9-16a,'-80. 
9.  a?2 +  13  a; +  30.  10.  x2-6x-55. 

11.  aj*  +  5a?,-14  12.  z4-17x2  +  72. 

13.  *4  +  10aj2  +  24.  14.  a4  +  6z2-27. 

15.  z6-5arJ-84.  16.  a8 +  8 a!4 -9. 

17.  ^  +  2/)2  +  2(a  +  2/)-35.  18.  (arJ+32/)2+13(ar2+3  7/)+42. 

19.  a2  +  4a&-12  62. 

Here  4  a&  is  the  product  of  the  square  root  of  the  first  term,  and  the  sum 
of  two  factors,  6  b  and  —2  b,  of  the  last  term.  Hence  the  factors  of 
a2  +  4  ab  -  12  b2  are  a  +  6  5  and  «  -  2  ft. 

20.  arJ-9a*?/-362/2.  21.   a^  +  llafy2  +  28#4. 
22.   x*-9x?y2  +  l$y*.  23.   z4  +  25 afy  +  100 ?/2. 
24.   ary  +  3a*/z2-2824.                25.  a;4  -  afyV  -  56  y4z\ 


62  HIGHER  ALGEBRA 


95.     Prob.     To  factor  a  trinomial  of  the  form  ax2m  +  bxm  +  c. 

Solution.      ax2m  +  bxm  +  c  =  -  [(aaf)2  +  6  (aaf)  +  ac],     since 

ft 

dividing  and  multiplying  by  the  same  quantity  does  not  change 

the  value  of  the  expression.     The  part  within  the  brackets  may 

now  be  factored  by  the  process  of  the  last  article  if  b  is  the  sum 

of  two  factors  of  ac. 

As  the  part  within  the  brackets  is  to  be  divided  by  a,  we  may 

divide  one  of  its  factors  by  the  whole  of  a,  or  one  of  them  by  one 

factor  of  a,  and  the  other  by  the  remaining  factor  of  a,  and  thus 

have  two  integral  binomial  factors  of  the  original  trinomial. 

EXAMPLES  XXIV 
Factor  the  following : 

1.  6x2  +  7x-20. 

Operation.        6  x2  +  7  x  -  20  =  J  [(6  x)2  +  7  (6  x)  -  120]. 

Here  7  (6x)  is  the  product  of  the  square  root  of  the  first  term  and  the  sum 
of  two  factors,  15  and  —  8,  of  the  last  term.  Hence  the  factors  of  the  part 
within  the  brackets  are  6x+15  and  6  x  —  8.  Dividing  the  first  of  these  by 
3,  one  of  the  factors  of  6,  and  the  other  by  2,  the  remaining  factor  of  6,  we 
have  for  the  factors  of  the  original  trinomial  2  x  +  5  and  3  x  —  4. 

2.  3x2-2«-5. 

Operation.      3  x2  -  2  x  -  5  =  \  [(3  x)2  -  2  (3  x)  -  15] 

=  i  (3  x  -  5)(3  x  +  3)  =  (3  x  -  5)(x  +  1). 

3.  12  x2  -5  z-2. 

Operation.     12  x2  -  5  x  -  2  =  T\  [(12  x)2  -  5  (12  x)  -  24] 

=  &  (12  x  -  8) (12  x  +  3)  =  (3  x  -  2) (4  x  +  1). 

4.  6£?  +  5a-4.  5.   10  a;2 -11  ^  +  3. 
6.  8 a?  +  26  a?  +  21.  7.  6  a?  +  13  a>  -  15. 

8.   12  a;2 -25  a +  12.  9.   15  a?  +  14  <e .-  & 

10.  6  £2  —  7  a;#  —  3  y2.  *. 
Operation.     6  x2  -  7  xy  -  3  */2  =  \  [(6  x)2  -  7  y  (6  x)  -  18  y2] 

=  K6x-9?/)(6x+2?/)  =  (2x-3?/)(3x+!/) 

11.  6  x2  -  19  a#  +  10  y2.  12.   10  «2  -  3  xy  -  18  f. 
13.   7  a4 +  25  a2 +  12.                     14.   14  a4  +  25  a2  +  6. 


FACTORING  63 

15.   (>a4  +  7aV/-3b\  16.   8  a6  +  14  a*b2  -  15  b\ 

17.   15  a6  -  23  arty2  -  28  y4.  la   o.t-8  +  11  a;4  -  12. 

19.   8^-6^-35.  20.   10  x10  -  31  a?Y  +  24  /'. 

96.  The  methods  of  Arts.  94  and  95  will  give  the  factors  of 
any  trinomial  of  the  form  ax2™  -f  bxmyn  -f-  cy2n  when  the  factors  are 
rational.  The  finding  of  such  factors  when  irrational  or  imagi- 
nary requires  the  solution  of  a  quadratic  equation. 

97.  A  polynomial  having  more  than  three  terms  may  sometimes 
be  factored  by  first  removing  monomial  factors  from  two  or  more 
groups  of  terms. 

Thus,  let  it  be  required  to  factor  am  +  an  +  bm  -f-  bn. 

am  -\-  an  -f  bm  +  bn  =  a  (*»  +  n)  -f  b  (m  +  n). 

We  now  see  that  the  binomial  factor  m  +  n  is  common  to  the 
two  terms,  being  contained  a  times  in  the  first  and  b  times  in  the 
second.     Hence  the  factors  are  m  -f  n  and  a  +  b. 

98.  Wlien  a  part  of  a  polynomial  can  be  factored  by  any  of  the 
preceding  methods,  we  may  sometimes  find  the  third  terms  of  trino- 
mial factors  as  follows : 

Let  it  be  required  to  find  the  factors  of 

6  x2  -  5  xy  -  2  x  +  43  y  -  2 1  y2  -  20 . 

This  has  the  form  of  the  product  of  two  trinomial  factors,  each 
having  terms  in  x  and  y,  and  a  term  containing  neither. 

The  factors  of  the  part  6  x2  —  5xy  —  21  y2  are,  by  Art.  95, 
Sx  —  7y  and  2x  +  3y. 

The  product  of  the  terms  not  containing  x  and  y  is  —  20.    Hence 

—  20  must  be  resolved  into  two  factors  such  that  the  sum  of  the 
products  obtained  by  multiplying  one  of  these  factors  by  3  x  and 
the  other  by  2  x  shall  be  —  2  x;  also  the  sum  of  the  products 
obtained  by  multiplying  one  of  these  factors  by  —  7  y  and  the 
other  by  3  y  shall  be  43  y.     By  trial  these  are  found  to  be  5  and 

—  4.     Hence 

6  x2  -  5  xy  -  2  x  +  43  y  -  21  f  -  20  =  (3  x  -  7  y  +  5)  (2  x  +  3  y  -  4) . 


64  HIGHER   ALGEBRA 

EXAMPLES   XXV 

Factor  the  following : 

1.  ax*  -2  ax2  -3  x  +  6. 

Operation,     ax3 -2  ax2-3x+6  =  ax2(x-2)  -3(x-2)  =  (x-2)(ax2-3). 

The  result  may  just  as  well  be  obtained  by  placing  the  first  and  third 
terms  in  one  group,  and  the  second  and  fourth  in  the  other. 

2.  x2  —  mx  —  nx-\-  mn.  3.   a?x  +  ax  -f  asy  +  ay. 

4.  5a^-20a^-2a;  +  8a.  5.   28  a2-21  a^  +  24a-18  2/. 

6.  xi  —  arfy  +  #z2  —  yz2.  7.   a&a£  +  bxy  —  axy  —  y2. 

8.  9  a2  +  6  ab  -  15  ac- 10  be. 

9.  3^-9i»y  +  5a;2-15^2-2^  +  62/2- 

10.  x2y2 -4;X2  +  3xy2-12x-  10 y2  +  40  (four  factors). 

11.  10  x>+  11  a#  - 14  xz  -  6  /  +  17#b  -  12  z2. 

Operation.  This  has  the  form  of  the  product  of  two  trinomials,  each 
having  terms  in  x,  y,  and  z. 

By  Art.  95,  the  factors  of  the  part  10  x2  +  11  xy  —  6  y2  are  5  x  —  2  y  and 
2x  +  Sy. 

The  product  of  the  terms  not  containing  x  and  y  is  —  12  z2.  Hence  —  12  z2 
must  be  resolved  into  two  factors  such  that  the  sum  of  the  products  obtained 
by  multiplying  one  of  these  factors  by  5  a;  and  the  other  by  2  x  shall  be 
—  14  xz  ;  also  the  sum  of  the  products  obtained  by  multiplying  one  of  these 
factors  by  —  2  y  and  the  other  by  3  y  shall  be  17  yz.  By  trial  these  are 
found  to  be  3 z  and  —  4 z.  Hence  the  factors  sought  are  5a;  —  2y  +  3z  and 
2x  +  3y-4z. 

12.  x2  +  3xy  +  5x  +  2y2  +  8y  +  6. 

13.  2  x2  -  5  xy  -  12  y2  -  9  xz  -  8  yz  +  4  z2. 

14.  6  x2  -  4  a?y  -  12  xz  +  9  a#  -  6  #2  -  5  yz  +  6  z2. 

15.  12  x2  4- 14  ^  -  11  x  -  10  y2  +  25  y  -  15. 

16.  6  a2  -  7  ^  -  3  ?/2  -  9  x  +  30  y  -  27. 

17.  4  x2  -  2  as  -  9  y2  -  27  yz  -  20  z2. 

18.  15  x4  +  sty*  -  19  a,-2z  -  6  2/4  +  19  y2z  -  10  z4. 


FACTORING  65 

99.  Theorem.  The  difference  of  any  two  quantities  is  a  divisor 
of  the  difference  of  the  same  powers  of  the  quantities. 

The  sum  of  two  quantities  is  a  divisor  of  the  difference  of  the 
same  even  powers,  and  the  sum  of  the  same  odd  powers  of  the 
quantities. 

Dem.  1st.  Let  x  and  y  be  any  quantities,  and  n  any  positive 
integer.     Then  x  —  y  divides  xn  —  yn. 

Supplying  the  missing  terms,  changing  the  sign  of  the  coeffi- 
cient of  the  second  term  of  the  divisor,  and  proceeding  with  the 
division  as  in  Art.  79,  we  have 

xn  +  0  xn~ly  +  0  xn~y  +  0  xn~y yn  [1 

111  0 

It  is  seen  that  the  quantity  to  be  added  to  the  last  term  is  1, 
giving  zero,  and,  consequently,  that  the  division  terminates,  the 
quotient  being 

x"-1  +  xn~2y  +  xn~3y2  +  af~y  •  •  •  -f-  yn~\  (1) 

2d.  Let  n  be  even.     Then  x  +  y  divides  xn  —  yn. 
Proceeding  as  before,  we  have 

fl^+G«^y+-Oaf-y+0ir-y-"  -yn  \  -1 

-i         i        -i 

It  is  seen  that  if  n  is  even,  the  quantity  to  be  added  to  the  last 
term  is  1,  giving  zero,  and,  consequently,  that  the  division  termi- 
nates, the  quotient  being 

xn-\  _  xn-2y  _|_  xn-3y2  _  ^n-y  . . .    _  yn-l  (2) 

3d.  Let  n  be  odd.     Then  x  +  y  divides  xn  +  yn. 
Proceeding  as  before,  we  have 

x«  +  0  afty  -f  0  af-y  +  0  xn~Y  •  ••  +  yn  |  — 1 
-1  1  -1 

It  is  seen  that  if  n  is  odd,  the  quantity  to  be  added  to  the  last 
term  is  —  1,  giving  zero,  and,  consequently,  that  the  division 
terminates,  the  quotient  being 

xn-l  _  xn-2y  +  xn-3y2  _  ^-y  ...    +  yn-l^ 

which  is  the  same  as  (2). 
downev's  alg.  — 6 


66  HIGHER   ALGEBRA 

100.  Sen.  When  the  sum  or  difference  of  the  same  powers  of 
two  quantities  is  given,  these  theorems  enable  us  to  determine 
whether  it  can  be  factored,  and,  if  so,  what  one  of  the  factors  is. 
Then  the  other  can  be  written  at  once  by  the  form  (1)  or  (2). 

EXAMPLES  XXVI 
Factor  the  following : 

1.  x?  +  y\ 

Operation.  Since  this  is  the  sum  of  the  same  odd  powers  of  two  quan- 
tities, it  is  divisible  by  the  sum  of  the  quantities.  Hence  x  +  y  is  one  factor, 
and  the  other,  as  given  in  form  (2)  above,  is  xx  —  x*y  +  x2y2  —  xy%  +  y*. 

2.  x*  —  y6. 

Operation'.       x?  —  y6  =  (x3  +  y3)  (x*  —  y%) 

=  (x  +  y)  O2  -  xy  +  y2)  (x  -  y)  (x2  +  xy  +  y2)  • 

3.  xG  +  f. 

Operation.       x«  +  y3  =  (z2)3  +  y*  =  (x2  +  y) [(z2)2  -  {x2)y  +  y2] 
=  (x2  +  y)(x*  -x2y  +  y2). 

4.  243 -a5. 

Operation.     243  -  a5  =  3*  -  as  =  (3  -  a)  (8«  +  33  a  +  32  a2  +  3  a3  +  a4) 
=  (3 -a)  (81  +  27  a  +  9a2 +  3  a3 -fa*). 

5.  x*  —  y4.  6.    af  —  y5. 

7.    a6 -64.  8.    27flf  +  y. 

9.   32^-1.  10.   27  a3  +  125  ?/6. 

11.   8^6- 216.  12.   a9  +  66. 

101.  Prob.  To  find  by  trial  the  binomial  factors  of  a  polynomial 
of  the  form  Axn  +  Bxn~l  +  Cxn~2  •••  +L,  or  of  a  homogeneous  poly- 
nomial of  the  form  Axn  -f  Bx1l~ly  +  Cxn~2y2  •  ••  +  7^/M. 

Solution.  The  short  method  of  successive  division,  given  in 
Art.  81,  affords  a  most  useful  method  of  factoring  such  expres- 
sions. Since  the  product  of  the  factors  of  a  polynomial  will 
produce  the  polynomial,  it  is  evident  from  the  process  of  multi- 
plication that  the  product  of  the  last  terms  of  the  factors  will 


FACTORING  67 

produce  the  last  term  of  the  polynomial.  Hence  in  rinding  by 
trial  these  factors  we  need  to  try  for  the  last  terms  only  those 
numbers  that  are  exact  divisors  of  the  last  term  of  the  polynomial. 
For  example,  let  it  be  required  to  rind  the  factors  of 

x5-  3 xA  -  5 x*  -\-15x2  +  ix  -  12. 

The  binomial  factors  of  the  first  degree  are  each  x  ±  some  num- 
ber which  is  a  factor  of  12.  A  trial  of  the  smaller  factors  of  12 
results  in  the  following : 


3x4-5.^ 

+  WX2 

+  4  x  - 

-12| 

1 

■2     -7 

8 

12 

0| 

2 

0     -7 

-6 

0 

I 

3 

3         2 

0 

1 

-1 

2         0 

1 

-2 

0 

It  will  be  remembered  that  this  is  a  short  process  of  dividing 
successively  by  x  —  1,  x  —  2,  x  —  3,  x  -+- 1,  and  x  -f-  2.  Hence, 
as  each  division  is  exact,  the  factors  are  x  —  1,  x  —  2,  x  —  3, 
x  +  1,  and  x  +  2. 

102.  Sch.  It  will  usually  be  found  expedient  to  find  the  posi- 
tive numbers  first,  as  above.  When  all  the  coefficients  of  any 
row  are  plus,  none  but  negative  numbers  need  be  tried,  as  the 
successive  additions  of  positive  products  to  positive  numbers 
could  not  produce  zero.  Before  writing  down  the  various  sums, 
it  is  best,  if  the  coefficients  are  not  too  large,  to  run  through 
mentally  with  a  factor  of  the  last  term,  and  see  whether  the  last 
addition  gives  zero. 

103.  Note.  If  the  student  has  forgotten  the  short  process  of 
division  as  given  in  Art.  79,  with  its  application  in  successive 
division  as  given  in  Art.  81,  he  should  go  back  and  thoroughly 
master  it,  as  it  is  of  frequent  and  important  use  in  subsequent 
parts  of  the  work. 


68  HIGHER  ALGEBRA 

EXAMPLES  XXVII 
Factor  the  following : 

1.  rf  -  x4  -  13  x3  +  13  x2  +  36  x  -  36. 

Operation 

&  -    x4  -  13  x3  +  13  x2  +  36  x  -  36 1 1 

0-13  0  36  0| 2 

2-9-18  0  | 3 

5  6  0  \-2 

3  0  |_-3 

0 
The  factors  are  x  -  1 ,  x  —  2,  x  —  3,  x  +  2,  and  x  +  3. 

2.  3  a*4  +  13  a?  -  117  a?  -  243. 

Operation 

3x4  +  13x3+    0x2-117x-243| 3 

22  66  81  0|  —  3 

13  27  0 

The  factors  are  x  -  3,  x  +  3,  and  3  x2  +  13  x  +  27. 

3.   ar5- 6  a;2 +  11  a; -6.  4.   ar*  +  5a;2  +  3a;-9. 

5.   ar*-  6  ar>  + 13  a;-  10.  6.   a6  +  8  x2  +  17  a?  +  10. 

7.   a?}-13ar2  +  49a;-45.  8.   ^-15^+74^-120. 

9.   a;4 +  2^-3^-4^  +  4.        JO.   x4-  10  ar3+  35  x2-  50  x  +  24. 

11.  aj4-2a^-25aj2+26a;+120.      12.   ^-4^  +  8^-8  a; -21. 

13.  a*-6rf  +  5a*  +  12x-60. 

14.  a?-a**-9a?  +  5a?  +  16aj-12. 

15.  aj5-9aJ4  +  25a?-15aj«-26aj  +  24. 

16.  ar5-4a!4-5a?  +  20a*  +  4aj-16. 

17.  2a4-9ar}  +  4a:2  +  21a;-18. 

18.  5ar5+7a;4-21ar}-lla;2  +  32a;-12. 

19.  3aJ5-2a*4-41a8  +  56aja-4a;  +  48. 

20.  a;6  -  14  a;4  +  49  a;2 -36. 

21.  a;6-4ari-3a;4  +  24arJ-10ar2-32a;  +  24. 

22.  a?7  +  5a^  +  6a?-6a;4-15arJ-3a;2  +  8a;  +  4. 


FACTORING  69 

EXAMPLES  XXVIII 
Factor  the  following : 

1.   7fg2y-28f2gy2  +  42fsgy.         2.   afy8  -  7  tfy4  +  12  xy5. 
3.    m4  —  n4.  4.    1  —  2  V#  +  a,*. 

5.   256  a4  +  544  a2  +  289.  6.   1-c3. 

7.   X2_a,_72.  8.   ye'-z4. 

9.   ^4-^-17^  +  15.  10.   z4-9.t2-90. 

u.  ?5_^.  +  i§.  12.  ^-2. 
m1     mx2      x4  b2     a2 

13.   125 +  64  a3.  14.   ^-15^  +  47^  +  63. 

15.   a2 +  23  a +  22.  16.   a?  +  b\ 

17.   cG-dc'.  18.   c~G-d-G. 

19.   1-13^  +  22^.  20.  4X2  +  8  x  +  3. 

21.   a0  -  b~6.  ■  22.   or2  +  6  xy  -  16  ?/2. 

23.   --6-10.  24.   507ra4+1326m2n*+867n3. 
x4 

25.   ^a4wi-22T«2w^2n+2+¥1T^n+4-  26-   3a  +  3  6-6Va6. 

27.   a5  +  &5.  28.   15  a  +  5  a#  —  x  —  3. 

29.   ^+3^-15  ^-19  a +  30.  30.   12aV*-12a2x*  +  3a2. 

31.  a:y-29^2  +  54. 

32.  21  abed  —  28  cda;?/  +  15  abmn  —  20  mnxy. 

33.   2z2-13a  +  6.  34.   3^-12^2-4?/2  +  l. 

35.   ^-14^+32^+95^+63.  36.   ar9-a-9900. 

37.   x2  +  ax  +  #  +  a.  38.   x4  —  11  x?y  +  18  y2. 

39.   6aj3-7ax2-20a2a;.  40.   ^-20^+30^+19^-30 

41.   x2"1  +  31  xm  -  32.  42.   af— a£  — 2#— 2. 


x10       ?/100 


«>«[!+g-2«* 


70  HIGHER  ALGEBRA 

45.  72  cdV  -  84  cd*m2  +  96  c2cl2m2.     46.   10  a4 +- 79  x2  -  8. 

47.  2x7y  +  54:xy4.  48.   6  <e°  +  19  afy2  —  7  ?/4. 

49.  x6-x-5-20x'4  +  50arJ-ll^2-49a,-H-30. 

50.  9  x2  -\-  4  y2  -\- z2  —  12  xy  -\- 6  xz  —  4  yz. 

51.  3»G-17^  +  23x4  +  9x3-14a;2-4a-24. 

52.  9  a2  +  3  xy  -  2  y2  -  3  a  +  13  y  -  20. 

53.  x2  +12  ay  — 14  as  +  36  ?/2  -  84  2/2  +  49  z2. 

54.  2  x7  -  3  a*  -  14  x5  +  20  ^  +  28  a*1  -  37  x2  -  16  a?  +  20. 

55.  10^  +  16a#  +  13a;z-8?/2-102/z-3z2. 

56.  Is  a:17  +  y17  divisible  by  x  +  y?  by  x  —  y ? 

57.  Is  Xs5  +  2/119  divisible  by  a*  -  y7  ?  by  ar5  +  /  ? 

58.  Write  by  form  (2),  Art.  99,  the  quotient  of  a10  +  bw  divided 
by  a2  +  b\ 


CHAPTER   VII 

HIGHEST  COMMON  DIVISOR  AND  LOWEST  COMMON 
MULTIPLE 

SECTION  I  — HIGHEST  COMMON  DIVISOR 

104.  The  Highest  Common  Divisor  of  two  or  more  algebraical 
quantities  is  the  quantity  of  highest  degree  that  will  exactly 
divide  each  of  them. 

The  abbreviation  h.  c.  d.  or  H.  C.  D.  is  often  used  for  highest 
common  divisor.  The  name  highest  common  factor,  with  the 
abbreviation  h.  c.  f.  or  H.  C.  F.,  is  also  used. 

105.  Quantities  are  said  to  be  prime  to  each  other  when  they 
have  no  common  factor  other  than  unity. 

106.  Note.  As  applied  to  literal  quantities,  highest  common 
divisor  is  preferred  to  greatest  common  divisor,  since,  when 
numerical  values  are  assigned,  the  quantity  having  the  higher 
degree  may  be  the  smaller. 

Thus,  for  positive  values,  if  a  >  1,  az  >  a  ;  but  if  a  <  1,  a%  <  a.  Again, 
for  values  of  x  and  y  greater  than  1,  if  x  >  y,  x2  —  y'2  >  x  —  y  ;  but  if  x  <  y, 

107.  Theorem.  The  highest  common  divisor  of  two  or  more  quan- 
tities is  the  product  of  their  common  prime  factors. 

This  follows  directly  from  the  definition. 

EXAMPLES  XXIX 
Find  the  highest  common  divisor  of  each  of  the  following: 
1.   72  aW,  84  a*b2c,  and  180  aW. 

Solution.  72  a4W  =  23  •  32  a*bW, 

84  a862c   =2».8  •  7  a?b% 

180  aWc*  =  22  •  32  •  5  a664c8. 

71 


72  HIGHER  ALGEBRA 

The  factors  of  highest  degree  common  to  all  are  22,  3,  a3,  62,  and  c. 
Hence  the  h.  c.  d.  is  the  product  of  these,  or  12  a?b2c. 

2.  48a264,  204  a?b2,  and  228  asbs. 

3.  81  oY*4,  123  aYs8,  and  315  aY- 

4.  6  m\x  —  ?/)3  and  9  m\x  —  yf. 

5.  aY  +  2  ar3?/2  and  aY  —  4  a;4?/2. 

6.  a;2  4-  6x  +  &  and  a?  4- 3  a;  4- 2. 

7.  a;2  +  x  —  6  and  x2  —  4. 

8.  a;2  —  1 ,  x2—  3x-\-2,  and  a;2 -f  6 a;  —  7. 

9.  ar3  4-1,  a*-l,  and  &'—2x  —  3. 

10.  2ar*-3a;-2  and  4x2  +  8^  +  3. 

11.  a2  +  2aH  &2,  a2 -  &2,  and  a3  4-  b3. 

12.  a?3  — a?,  x3 4- 9^2— 10a?,  and  xP  —  x. 

13.  ar*  4-  3  afy  4-  2  ar#2  and  a;4  4-  6  ar3?/  4-  8  a;2?/2. 

14.  ^_x_42,  x2-4b-60,  and  ar2 4- 12^4-36. 

15.  2ar2-7a;4-3  and  3b2-7b-6. 

16.  a;4-2ar3-13ar2+38a;-24  and  ^-4^-7^  +  34^-24. 

Operation.     We  proceed  as  follows,  by  the  process  of  Art.  101 : 

X4  _  2  x3  -  13  x2  +  38  x  -  24  j_J_  x*  -  4  a3  -  7  a2  +  34  x  -  24  j     1 

-  1       -  14  24  0  ) 2  -3-10        24  0  |     2 

1-12  0  13  _1_12  0  14 


4  0  [-4  3  0  [-3 

0  0 

The  common  factors  are  thus  seen  to  be  x  —  1  and  x  —  2.     Hence,  multi 
plying  by  the  process  of  Art.  60,  the  h.  c.  d.  is  x2  —  3  x  +  2. 

17.  x3  4-  2  sc2  4-  a;  4-  2  and  ar4  —  4  as8  —  b  —  2. 

18.  a;3  4-  4  x2  -  8  a;  4-  24  and  x4  -  x3  4-  8  x  -  8. 


HI0HB8T  COMMON  DIVISOR  73 

19.  2ar3  +  x2  -  x  -  2  and  6ar3  -  4ar>  +  2a;  -  4. 

20.  x4-5xi-\-5x2-x-12  and  a;4  -  2x>-  12a?  +  11  a;  +  20. 

21.  ar3- 13a; +  12  and  a4  +  3 or3  +  12 x  -  16. 

22.  ^-4^-7 ^+34 z-24  and  a^-G^+ar3 +36^-20^-48. 

23.  a^-1  and  x4  +  ar3-  9a,-2  +  10a:  -  8. 

24.  12  x4  -  24  afy  +  12  ar2?/2  and  8  x*y2  -  24 afy8  +  24  xy4  -8y>. 

25.  a4  -  asb  -  a2b2  -2  b4  and  3  a3  -  7  a2b  +  3  a£2  -  2  63. 

26.  a4-5a36  +  5a2fr2-a&3-12&4  and  a4  -  2  a36  -  12  a2b2  + 
lla&3  +  20  64. 

27.  3ar,  +  2a;4-47ar3  +  10ar2  +  128a;-96  and  3  ar5- 10  a;4- 31  ar3 
+  94  ar  +  16ar-96. 

28.  ar5-5a;4-15ar3  +  65ar2  +  74a;-120  and  V  -  4  x4  -  16  a^ 
+  46^4-  63  a;  -90. 

Operation.  By  the  process  of  Art.  101,  the  common  factors  are  found  to 
be  x  —  5,  x  +  3,  at  +  2,  and  a  —  1.  After  writing  by  Art.  60,  the  product 
of  the  first  two,  we  proceed  as  follows,  by  the  process  of  Art.  59 : 

(2)       1       -2       -15 
(-1)  0       -19       -30 

_1       _io       _n      30 

Hence  the  h.  c.  d.  is  a*  -  xs  -  19  x2  -  11  x  +  30. 

29.  x*-5x4  +  7ar3-7ar2  +  16a;-12  and  x5 - 8 x4  +  26 x* - 46 ar2 
+  45  a;- 18. 

30.  ar5  +  2ar4-15ar3-8ar2  +  68a;-48  and  ar5  +  8  x4  +  15  ar3  - 
20a;2 -76  a; -48. 

.  31.   x*  +  6x*  +  6x*  -16a,-2  -15a;  +  18    and   a,-5  +  2 a;4  -  10 a,-3  - 
Sx2  +  33a; -18. 

32.  3ar5+a;4-llx3+3arJ+8a,'-4  and  3a,-5 +7 x4 -3 ar'-llarM- 4. 

33.  2  y;  +  5.r5-  6a;4  -4ar3  +  2a;2  -29a;  +  30  and  2ar5-7a;4  + 
4ar3-4ar2  -2a;  +  15. 


74  HIGHER  ALGEBRA 

108.  When  one  of  the  polynomials  is  readily  factored  and  the 
other  not,  ice  may  find  by  tried  what  factors  of  the  first  are  contained 
in  the  second,  and  thus  obtain  the  h.  c.  d,  of  the  two. 

EXAMPLES  XXX 
Find  the  h.  c.  d.  of  each  of  the  following: 

1.  3*+  if  and  5x4  -  18afy  +  I2xhf  -7 xif  -  0>y\ 

The  factors  of  x*  +  ?/3  (form  (2)  of  Art.  99)  are  x  +  y  and  x2  -  xy  +  y*. 
By  trial  the  second  of  these  is  found  to  be  a  factor  of  the  second  polynomial. 

2.  x*  —  1  and  2  ar3  —  x2  —  x  —  3. 

3.  9«*-16  and  9^-15^  +  10^-8. 

4.  x3  +  2  afy  +  4  #?/2  +  3  if  and  a*1  +  afy  +  4  a,-2?/2  +  .ry3  +  3  y\ 

5.  a^  +  T/5  and  7  x6  -  10x*y  +  lOtfy2  -10x2f  +  10xyA  -  3f. 

109.  Note.  When  the  polynomials,  or  one  of  them,  can  be 
readily  factored,  the  processes  of  Arts.  107  and  108  of  finding  the 
h.  c.  d.  are  the  most  expeditious.  Otherwise,  the  process  of  Art. 
Ill,  similar  to  that  used  for  numbers  in  Arithmetic,  may  be 
employed. 

110.  Theorem.  1st.  A  divisor  of  a  quantity  is  a  divisor  of  any 
multiple  of  that  quantity. 

2d.  A  common  divisor  of  two  quantities  is  a  divisor  of  their  sum 
and  also  of  their  difference. 

The  first  is  self-evident. 

For  the  second,  let  d  be  a  common  divisor  of  a  and  b,  and 
q  and  q'  the  respective  quotients.     Then 

\a  =  qd, 

and  b  =  q'd,   ■ 

whence  a±b  =  qd  ±  q'd  =  (q  ±q')d. 

Hence  d  is  a  divisor  of  a  ±  b. 


HIGHEST  COMMON  DIVISOR  75 

111.  Prob.  To  find  the  h.  c.  d.  of  two  polynomial*  which  are  not 
readily  resolved  into  their  prime  factors. 

Rule.  Having  arranged  the  polynomials  with  reference  to  the 
same  letter,  remove,  and  reserve  as  factors  of  the  h.  c.  d.,  any  mo- 
nomial factors  common  to  both  polynomials,  and  reject  from  each 
polynomial  all  other  monomial  factors. 

Divide  the  reduced  polyyiomial  of  higher  degree  in  the  letter  of 
arrangement  by  the  other  (if  both  are  of  the  same  degree,  either  may 
be  used  as  the  dividend),  first  multiplying  the  dividend,  if  necessary, 
by  any  number  that  will  avoid  a  fraction  in  the  quotient. 

Reject  from  the  remainder  any  monomial  factors  and  divide 
the  former  divisor  by  this  reduced  remainder,  first  multiplying } 
if  necessary,  by  any  number  that  will  avoid  a  fraction  in  the 
quotient. 

Continue  the  process  either  until  there  is  no  remainder,  or  until 
the  letter  of  arrangement  disappears  from  the  remainder.  In  the 
latter  case  the  two  reduced  polynomials  are  prime  to  each  other,  and 
in  the  former  case  the  product  of  the  last  divisor  and  the  reserved 
common  monomial  factors  ivill  be  the  h.  c.  d.  of  the  given  polynomials. 

Dem.  Rejecting  from  or  introducing  into  either  polynomial 
or  any  remainder  a  monomial  factor  will  not  affect  the  h.  c.  d., 
since  the  h.  c.  d.  is  composed  of  those  factors  only  that  are  common 
to  the  two  polynomials. 

If  A  and  B  represent  the  reduced  polynomials,  q,  q',  etc.,  the 
successive  quotients,  and  R,  R',  etc.,  the  successive  remainders, 
and  if  we  suppose  the  third  remainder  to  be 
0,  the  work  (except  rejecting  and  introducing      B)A(q 
factors)  will  stand  as  in  the  margin.  qll 

A  divisor  of  A  and  B  is  (Art.  110)  a  divisor  ft)B(q' 

of  A  —  qB,  or  R ;  and  a  divisor  of  B  and  R  is  q'R 

a  divisor  of  B  -  q'R,  or  R'.      Therefore    any  .  R^Riq" 

divisor  of  A  and  B  is  a  divisor  of  R';    and  q"R' 

since  R'  is  its  own  highest  divisor,  it  is  the  0 

highest  common  divisor  of  A  and  B.      Hence 
the  product  of  R'  (the  last  divisor),  and  the  reserved  common 
monomial  factors  is  the  h.  c.  d.  of  the  given  polynomials. 


76  TI1GUER  ALGEBRA 

EXAMPLES   XXXI 
Find  the  h.  c.  d.  of  the  following : 

1.  8^-48afy  +  84afy2-40ary  and 

6a%  -  48afy2  +  126  ofy3  -  120  afy4  +  24  xy5. 
Operation 

Removing  4  x2  from  the  first  and  6  xy  from  the  second,  reserving  2  ac,  the 
h.  c.  d.  of  these,  as  a  part  of  the  h.  c.  d.  of  the  polynomials,  we  have  left 

2  %*  -  12  x2y  +  21  xy2  -  10  #3   and    x4  -  8  x3y  4-  21  x2?/2  -  20  x?/3  +  4  y2. 

Multiplying  the  second  by  2  to  avoid  fractions,  and  using  Synthetic 
Division  (in  which  the  signs  of  all  the  terms  of  the  divisor  except  the  first 
must  be  changed),  we  have 

2  x4  -  16  x*y  +  42  x2y2  -  40  xy*  +  8  y*  |  2  x3  +  12  x2y  -  21  xy2  +  10  y3 
12        -21  10        -20        x-2y 

-4        -_24  42 

-    3  12        -  12 

.'.  the  remainder  is  —  3  x2  +  12  xy  —  12  ?/'2. 

Rejecting  —  3,  we  have  for  the  new  divisor  x2  —  4  xy  +  4  y2. 

2x3  -  12x2*/  +  21  xy2  -  10y3  [  x2  +  4xy  -  4y2 
_8       -    8  16        2x-4y 

-    4        -10 

-    3  ~6 

.-.  the  remainder  is  —3x  +  6y. 

Rejecting  —  3,  we  have  for  the  next  divisor  x  —  2y. 

x2-4xy  +  4y2\_2 
-2  0 

Hence,  in  dividing  by  x  —  2  y  there  is  no  remainder.  Multiplying  this 
last  divisor  by  the  common  factor  2x  reserved  at  the  beginning,  we  have  for 
the  h.  c.  d.  of  the  original  polynomials  2  x  (x  —  2  y)  =  2  x2  —  4  xy. 

2.  3^  +  9afy-6aW2-62/3  and  24 a?  +  6 x2y  -  12 xy2  -  18 f. 

3.  21a?-32«8-54a>-7  and  21^-4^-15^-2. 

4.  10^  +  ^-9^  +  24  and  20a4-  17.r2  +  48a-3. 

5.  2a!4-7<B3  +  ll<B2-8a;  +  2  and  2  a;4  +  a?  -  9^2  +  Sx  -  2. 

6.  4^+14x-4+20a;3+70a;2  and  8 af+28^- 8  x5- 12  a,*4 +56 a?. 


LOWEST  COMMON  MULTIPLE  77 

112.  Prob.     To  find  the  h.  c.  d.  of  three  or  more  polynomials. 
Rule.     Find  the  h.  c.  d.  of  any  two  of  the  given  polynomials,  then 

find  the  h.  c.  d.  of  this  h.  c.  d.  and  a  third  polynomial,  and  so  on 
until  all  have  been  used. 

Dem.  If  A,  B,  C,  etc.,  represent  the  given  polynomials,  and 
M  the  h.  c.  d.  of  A  and  B,  M  contains  all  the  factors  common  to 
A  and  B  (Art.  107).  If  N  represent  the  h.  c.  d.  of  M  and  C,  N 
contains  all  the  factors  common  to  M  and  C,  and  consequently 
all  the  factors  common  to  A,  B,  and  (7;  and  so  on. 

EXAMPLES  XXXn 
Find  the  h.  c.  d.  of  the  following : 

1.  x^+llx  +  ZO,  2a2  +  21;c  +  54,  and  9ar3  +  53x?  -  9x  -  18. 

2.  x^-la?  +  5x2  +  31  x  -30,  x4  -  9  ar3  +  21  x2  +  x  -  30,  and 
x4-8xi  +  6x2  +  72x-135. 

3.  10ar5  +  10ary  +  20afy,  20^  +  2^,  and  4 y*  +  12 x-y  +  4 ar3?/ 
+  12ar/. 

SECTION  II  — LOWEST  COMMON  MULTIPLE 

113.  The  Lowest  Common  Multiple  of  two  or  more  algebraical 
quantities  is  the  quantity  of  lowest  degree  that  is  exactly  divis- 
ible by  each  of  them. 

The  abbreviation  1.  c.  m.  or  L.  C.  M.  is  often  used  for  lowest 
common  multiple. 

114.  Prob.  To  find  the  1.  c.  m.  of  two  or  more  algebraical 
quantities. 

Rule.  Multiply  one  of  the  quantities  (the  one  of  highest  degree 
when  they  differ  in  degree)  by  all  the  factors  of  the  others  that  are 
not  found  in  it. 

Dem.  Let  the  quantities  be  represented  by  A,  B,  C,  etc.  As 
any  one  of  the  quantities,  A  for  example,  is  its  own  lowest 
multiple,  the  1.  c.  m.  of  all  the  quantities  must  contain  it  as  a 
factor.  As  the  1.  c.  m.  contains  also  B,  C,  etc.,  it  must  contain  all 
the  factors  of  these  quantities ;  hence  any  of  these  factors  that 
are  not  found  in  A  must  be  introduced. 


78  HIGHER  ALGEBRA 

115.  Soh.  In  applying  this  rule,  if  the  factors  of  the  poly- 
nomials are  not  readily  found,  we  may  find  the  h.  c.  d.  in  the  usual 
way,  and  then  find  the  other  factors  by  dividing  each  polynomial 
by  the  h.  c.  d. 

EXAMPLES   XXXIII 
Find  the  1.  c.  m.  of  the  following : 

1.  x3-2x2-5x  +  6,  ^-3 ^-z  +  3,  and  ar?  +  4a2  +  *— $ 
Solution.    Factoring  by  the  method  of  Art.  101,  we  have 

x*-2x*-bx  +  Q=(x-  1)0-3)0  +  2), 
xs -Sx2-  £  +  3  =0-1)0  ~3)0  +  1). 
x*  +  4x2+    x-6=(x-l)(x  +  2)(x  +  3). 

The  second  and  third  polynomials  contain  one  factor  each  not  found  in 
the  first.  Hence  multiplying  the  first  polynomial  by  these  factors,  using  the 
method  of  Art.  59,  we  have  for  the  1.  c.  m. 

03-2x2-  5x  +  6)0  +  l)0  +  3)=:«5  +  2ic4-  10x3-20x2  +  9x+  18. 

2.  x2  +  2,  x2  -  2,  and  a;4  -  4. 

3.  x2-y2,  (x  +  y)2,  (x  -  y)2,  and  (x  +  yf. 

4.  x>-9  and  a2  +  10  a -f  21. 

5.  x2-2x-lo  and  a2 -4a -21. 

6.  £2-3#-70  and  a)3-39#  +  70. 

7.  x2  -  9,  x2  +  x  -  12,  and  x2  +  2  x  -  15. 

8.  x2  +  x  —  2,  a2  —  x  —  6,  and  a,*2  —  4  #  +-  3. 

9.  2^-7^-4,  4a2  +  10a; +  4,  and  6x?  -1  x  -5. 

10.  a3  -  2  a2  -  5  a  +  6,  a3  -  3  a2  -  a  +-  3,  and  a3  +  4  a2  +  a-6. 

11.  ar3-2a;22/+4a!?/2-8^,  ar}-r-2afy+4a;?/2+8  2/3,  and  ar-4?/2. 

12.  a^  +  Sa^  +  Sa^-Sz-e,  ^  +  6^  +  11^4-6,  and  f  +  4f 
+  a;-6. 

13.  a4  -  1,  a3  +  a2  +  a  +  1,  a3  —  a2  +  a  —  1,  and  a2  +  1. 

14.  a4-10a;2  +  9,  a;4  +  10 ar> +- 20 a;2 -10 a; -21,  anda4  +  4ar3 
-22a2  -4a;  +  21. 

15.  6o2-5a*/  +  2a;-62/2  +  232/-20  and  8  x2  -  14  a^  +  22  x 
+  3/ -13  2/ +  12. 


CHAPTER   VIII 
FRACTIONS 

116.  An  Algebraic  Fraction  is  an  indicated  operation  in  division 

711 

when  written  in  the  form  —  or  m/n. 

n 

In  this  form  the  dividend,  written  above  or  before  the  vinculum, 

is  called  the  Numerator,  the  divisor  is  called  the  Denominator,  and 

the  quotient  is  called  the  Value  of  the  Fraction. 

117.  When  the  denominator  is  a  positive  integer,  it  indicates, 
as  in  case  of  arithmetical  fractions,  the  number  of  equal  parts 
into  which  a  unit  is  conceived  to  be  divided,  and  the  numerator 
indicates  the  number  of  these  parts  taken ;  but  it  would  be  absurd 
to  represent  a  unit  as  divided  into,  say  5|  equal  parts,  or  into  —  4 

equal  parts.     Now  in  such  a  fraction  as  — ,  m  and  n  are  un  re- 
ft 

stricted  in  value,  and  are  not  necessarily  positive  integers.  Hence 
the  denominator  of  an  algebraic  fraction  does  not  necessarily  indi- 
cate into  how  many  equal  parts  a  unit  is  conceived  to  be  divided. 

118.  A  quantity  is  said  to  have  the  Integral  Form  when  it  has 
no  part  in  the  fractional  form. 

Thus,  3  mn,  2x  —  3y,  ax3  -  bx2y  -f  cy3,  are  in  the  integral  form. 

119.  A  quantity  is  said  to  have  the  Mixed  Form  when  it  con- 
tains terms  in  both  the  integral  and  the  fractional  form. 

Thus,  a  +  — ,  x2  -  y  +  x  ~  y  ,  have  the  mixed  form. 
n  x2  —  y 

120.  A  Proper  Fraction  is  one  which  cannot,  without  the  use  of 
negative  exponents,  be  reduced  to  the  integral  or  mixed  form. 
If  its  numerator  and  denominator  contain  a  common  letter,  the 
numerator  is  of  lower  degree  in  this  letter  than  the  denominator. 

79 


80  HIGHER   ALGEBRA 

Thus,  x  +  y ,    — x~  +  ° — ,  are  proper  fractions. 

121.    An  Improper  Fraction  is  one  whose  numerator  is  not  of 
lower  decree  in  a  common  letter  than  its  denominator. 


Thus, 


x+5/      ,  1     \     z3  +  3a;2-4z  +  5/     ..  ,  -  ,      Sx  -10 


£(=1— JLA    ^  +  3^-4^  +  5/ 
6\  x  +  <$)  z2-2x  +  3       V 


K  +  6\  x  +  ti/  x2-2x  +  S       \  x2-2x  +  3 

are  improper  fractions. 

122.  A  Simple  Fraction  is  a  fraction  whose  numerator  and 
denominator  are  both  in  the  integral  form. 

123.  A  Complex  Fraction  is  a  fraction  having  its  numerator  or 
its  denominator  or  both  in  the  fractional  or  mixed  form. 

124.  Reduction  is  the  operation  of  changing  the  form  of  a  quan- 
tity without  changing  its  value. 

125.  The  Lowest  Common  Denominator  of  several  fractions  is  the 
lowest  common  multiple  (Art.  113)  of  their  denominators. 

,    126.    A  fraction  is  in  its  Lowest  Terms  when  its  numerator  and 
denominator  are  prime  to  each  other. 

REDUCTION  OF   FRACTIONS 

127.  Theorem.  Multiplying  or  dividing  both  numerator  and 
denominator  of  a  fraction  by  the  same  quantity  does  not  change 
the  value  of  the  fraction. 

Dem.  Represent  numerator,  denominator,  and  quotient  by  », 
d,  and  q,  respectively. 

Since  divisor  times  quotient  equals  dividend  (Art.  66),  we  have 

d  x  q  =  n. 

Since  multiplying  or  dividing  a  factor  of  a  product  multiplies 
or  divides  the  product,  multiplying  or  dividing  d  and  n  in  the 
above  equation  by  the  same  quantity  will  not  change  q. 

128.  Cor.  Changing  the  signs  of  all  the  terms  in  a  fraction  does 
not  change  the  value  of  the  fraction. 

For  this  is  equivalent  to  multiplying  or  dividing  both  numera- 
tor and  denominator  by  —  1. 


FRACTIONS  81 

129.  Prob.      To  reduce  a  fraction  to  its  lowest  terms. 

Rule.  Reject  from  numerator  and  denominator  all  factors  com- 
mon to  both;  or,  what  is  the  same  thing,  reject  from  both  their 
h.  c.  d. 

This  follows  from  Arts.  127  and  107. 

130.  Note.  The  student  should  endeavor  to  find  the  factors 
of  numerator  and  denominator,  and  after  finding  them  he  should 
cancel  those  that  are  common,  avoiding  whenever  possible  the 
laborious  method  of  Art.  Ill  of  finding  the  h.  c.  d.  In  none  of 
the  following  is  it  necessary  to  resort  to  this  method.  In  the 
last  seven  he  should  factor  by  the  method  of  Art.  101. 

EXAMPLES   XXXIV 

Reduce  the  following  to  their  lowest  terms : 

60  a&V  420  a^c2" 

84  abW  "    630  ab2cn' 

3    210  ahk*n  4  bx  +  x2 

'   330  aWd*  '  db  +  ax 

x2-^  6  3  a*b  -  6  a2b2 

5    X4  _  y4  '  4a2&2-8ttfc3' 

x2  _l_  7  x  _|_  10  8    x2-  x  -12 

"    aj»  +  4s-5"  '  tf  +  x-20 

±x*-4:X  +  l  10     v?  +  5  x  -  14 

'   4ar3-3a  +  l'  '  x*  +  10  a +  21 

ii  a+*Y-  i2.     a3+^    ■ 

'   (1  —  ic2)2  a2-\-2ax  +  x2 

ar»-27  14    a2  -  7  x  +  10, 

'   x*-2x-£  tf-Sx  +  G 

x2-^2  16     l-5a  +  6a2, 

'   (x  +  xyf  '   l-7a  +  12a2 

afr-afl*  18    (a +  &)'-(?. 

downey's  alg.  — 6 


82  HIGHER  ALGEBRA 

19    (a  +  bf  -  (c  +  df  2Q    6^-5^-4 

(a  +  c)2  -  (b  +  d)2  '    6  a;2  +  a; -12* 

^      (fr  +  c  +  d)2-ft2  22  s»-7s  +  6 

(a  -  &)2  -  (c  +  d)2  '   tf-  6  ar*  + 11  a: -6* 

Scg.     Factor  by  the  process  of  Art.  101. 

ar3  -  8  x2  +  19  x  -  12 


23. 


24. 


25. 


26. 


27. 


28. 


ar*  -  10  ar>  +  29  2  -  20 

2  a;4  -  jg  -  9  x*  +  13  g  -  5 

7  x4  -  26  x*  +  36  jb2  -  22  a;  +  5* 

gg  -  8  ar*  +  21  x  -  18 

3x4-22arJ  +  53a2-42a/ 

a4  +  10  ar»  +  35  ar*  +  50  a-  +  24 
ic4  +  a? -19  a?2-  49  a;-  30 

2  or5  -  9  a;4  +  8  a^  +  15  x2  -  28  g  +  12 
3  Xs  -  19  x4  +  45  a?  -  49  x2  +  24  a;  -  4* 

6  ar*  -  19  a;4  +  6  ar*  +  36  ar*  -  44  a;  +  15 

8  ic5  -  18  x4  -  3  ar3  +  37  a^  -  33  x  +  9  ' 


131.  Prob.  To  reduce  an  improper  fraction  to  an  integral  or 
mixed  form. 

Rule.  Divide  the  numerator  by  the  denominator,  continuing  the 
division  either  until  it  terminates,  or  until  the  remainder  is  of  lower 
degree  than  the  denominator. 

EXAMPLES  XXXV 

Reduce  the  following  to  integral  or  mixed  forms: 

1    6a?-4sg-18a?-10 
2s8-4s-l 

Operation.     Dividing  by  the  method  of  Art.  76,  we  have 

G  x3  -  4  x2  -  18 x  -  10  I  2a2  +  4s-f  1 
12  3  4    .    8x  +  4 

8  16 

1       -6 

Hence  the  result  is  3  x  +  4  -j g 

2  se'2  —  4  x  —  1 


4    ^ x  J  "*"       J  ~        9  5. 


2af- 

■  4  a2?/  4-  4  #?/2 

-25  2/ 

a»  —  3jf 

12  x3 

-8^  +  4a;- 

-5 

4ar  +  3 

2  a3- 

-  3  ar9  -  5 

FRACTIONS  83 

2    6  ar9- 12  a; -2  3    3  a3-  8  or0  +  6  a  +  3 

3x  '  x- 2 

Sue     In  such  examples  as  the  3d  and  4th  employ  the  method  of  Art.  101. 

m2  4-  n2  4-  2  mn  —  x  —  y 
m  +  n 

32  ar8  +  243 
2a?4-3 

8^+16^-10^-28^4-11 
x2  —  x—1  2xr  +  x  —  3 

10  a4+7  ar*-ll  arM- 13  x-  3     „    %4+4  afy 4-6 afy2 -f  4  ay3 4-  ?/4 
'  ~     2o34-7ari4-5a;-2  Xs +  3  x*y  +  3xy2  +  f 

132.  Prob.     To  reduce  a  mixed  quantity  to  the  fractional  form. 

Rule.  Multiply  the  integral  part  by  the  denominator  of  the 
fractional  part,  to  the  product  add  the  numerator  of  the  fractional 
part,  and,  place  the  sum  over  the  denominator  of  the  fractional  part. 

Dem.  The  value  of  the  integral  part  is  not  changed  by  multi- 
plying it  by  the  denominator  of  the  fractional  part  and  then 
indicating  its  division  by  the  same  quantity.  The  two  numera- 
tors may  now  be  added  and  written  over  the  common  denomi- 
nator, since  the  quotient  of  the  sum  is  equal  to  the  sum  of  the 
quotients,  the  divisor  being  the  same  (Art.  72). 

133.  Cor.  An  integer  may  be  reduced  to  the  form  of  a  fraction 
with  any  denominator  by  multiplying  it  by  that  denominator  and 
then  indicating  its  division  by  the  same  quantity. 


EXAMPLES  XXXVI 

Reduce  the  following  to  the  fractional  form : 

a  3x  —  4: 

3.   a-x-(a  +  *y.  4.   *-*„  +  ?-£*-. 

a—x  %+y 


84  HIGHER   ALGEBRA 


5.   a3  +  a26  +  a&2  +  &3  +  -^-.         6.    1  ■ 


a  —  b  2  be 

7.   S.2+     ll*  +  22    .  8    0_6      3a»ft  +  2ay-y 

or2  +  7  a?  +  10  a2  -  b2 

134.  Prob.  To  rreduce  fractions  having  different  denominators  to 
equivalent  fractions  having  the  lowest  common  denominator. 

Rule.  Divide  the  1.  c.  m.  of  all  the  denominators  by  the  denomi- 
nator of  each  fraction,  and  multiply  its  numerator  by  the  quotient. 

Dem.  The  object  of  the  division  is  merely  to  find  the  factor 
by  which  the  denominator  of  the  fraction  must  be  multiplied  to 
produce  the  lowest  common  denominator.  As  the  numerator  is 
multiplied  by  the  same  factor,  the  value  of  the  fraction  is  not 
changed. 

135.  Cor.  When  the  denominators  have  no  common  factors,  the 
lowest  common  denominator  is  the  product  of  all  the  denominators, 
and  each  numerator  is  multiplied  by  the  denomiyiators  of  all  the 
other  fractions. 

EXAMPLES  XXXVII 

Reduce  the  following  to  equivalent  fractions  having  the  lowest 
common  denominator : 

-        x  y  z  n         5  1  2 


a  +  b     a-b     a2-b2  2-2ar*     1-x2    3#-3 


x  x2 


x  y 


x  +  y     x*-y2     x*-y4  a2-b2     (a  +  b)2     (a-b)* 

K  2x  Sy         4z         .        1  x  x2 


x2  —  xy  +  y2    x3  +  y3    x  +  y        '   x  —  y     (x  —  y)2     (x  —  yf 

1  2  3 

tf  +  x  —  ti    ^4- 5  a +  6*    a.-2  — 4' 

11  1 


x  +  2    ar*  +  4a,-  +  4    x*+6x* +12x  +  4 

9    (a?-2)(a?Hrl)     (s+2)(a?-l)     (a?+2)(a?+l)     (x-2)(x-l) 
x?+x-2  xF-x-2  ^-3^+2  '      a?+3x+2 


ADDITION  AND   SUBTRACTION   OF  FRACTIONS         85 

ADDITION   AND   SUBTRACTION   OF   FRACTIONS 

136.    Prob.      To  add  or  subtract  fractions. 

Rule.  Reduce  the  fractions  to  equivalent  fractioyis  having 
the  lowest  common  denominator,  write  the  sum  or  difference  of  the 
numerators  over  the  lowest  common  denominator,  and  reduce  the 
result  to  its  simplest  form. 

Dem.  After  the  first  step,  which  does  not  change  the  values 
of  the  fractions  (Art.  134),  we  apply  the  principle  that  the  sum 
or  difference  of  the  quotients  is  equal  to  the  quotient  of  the  sum 
or  difference,  the  divisors  being  the  same  (Art.  72). 

EXAMPLES   XXXVIII 

Perform  the  operations  indicated  in  the  following : 

a2  +  x2        a2x    _  x3  -f-  a4x 
x  +  a2      x  —  a2      x2  —  a4 

Operation.  Reducing  to  equivalent  fractions  having  a  common  denomi- 
nator, we  have 

x3  -  a2x2  +  a2x  -  a*     a2x2  +  a*x  _  sc8  4  a*x  _  a2x  -  a4  _  a2(x  -  a2)  _     a8 
x'2  -  a4  x2  -  a*         x2  -  a*  ~  x2  —  a4         x2  —  a4        x  +  a2 

1      +-L-  3         *      +       * 


x  -f  y     x  —  y  x  -\-  x2     x  —  x2 

a2  +  ab  +  b2     a2-ab+b2        _    m2x  +  x*-x* 

4.    » •         5. 


a  4-6  a  —  b  4  x2  —  4  aj8  +  a;4     2  #  —  ar* 

6-     2         ..+  3 


ar*  +  a?  4-  a?  4- 1      ar*  —  sc2  4-  a;  —  1 

%  —  ct      a2  4-  3  ax     x  +  a 
a?  4-  a        a2  —  x2       x  —  a 

.      1      ,  1  2 


a;  —  3      a2  —  5  a!  -(-  6      a^  —  6  #  +  8 

i    ,       7       +»'+-s; 


#4-3      a?2  —  x  —  12      a;  —  4 
10    3a^-8        bx  +  1  2 


a^_l       aj84-aJ  +  l*-l 


86  HIGHER  ALGEBRA 

x  —  3  x  —  2  x  —  1 


11. 


12. 


a2  —  3  x  +  2      x2  —  4  a*  +  3     a;2  —  5  a;  +  6 

10a;2  "2 1 

(1  +  a2)  (1  -  4  x2)      1  +  a*      1  -  2  a' 


13    ft +  2  1  3  2 


14. 

a2  —  1              x  4-  1              a;  —  1 

tf4  +  ^  +  1        SB2  —  X  -f  1        it*2  +  3  +  1 

15 

11            a-(a-2  +  3) 

x*  _  1      (a-  +  1)«       (&  _  l)^       (x. ._  iy 

16. 

1                     05  +  1                X2  +  X  +  lt 

a  +  3     3*-3flS  +  ?        ^  +  27 
17         x2-2x-15  a»2+7a-+12  z2-16 


ar3-5a*2-9a+45      ar'+3a-2-9a;-27      ar{-4a-2-9a-+36 

MULTIPLICATION   OF   FRACTIONS 

137.  Theorem.  1st.  Multiplying  the  numerator  of  a  fraction 
multiplies  the  fraction. 

2d.  Dividing  the  numerator  of  a  fraction  divides  the  fraction. 
3d.  Multiplying  the  denominator  of  a  fraction  divides  the  fraction. 
4th.  Dividing  the  denominator  of  a  fraction  multiplies  the  fraction. 

Dem.  The  numerator  is  dividend,  or  quantity  to  be  measured, 
and  the  denominator  is  divisor,  or  measure.  Now  multiplying 
the  quantity  to  be  measured  multiplies  the  number  of  times  it 
contains  the  measure,  and  dividing  the  quantity  to  be  measured 
divides  the  number  of  times  it  contains  the  measure. 

Likewise,  multiplying  the  measure  divides  the  number  of  times 
it  is  contained  in  the  quantity  to  be  measured,  and  dividing  the 
measure  multiplies  the  number  of  times  it  is  contained  in  the 
quantity  to  be  measured. 

138.  Prob.      To  multiply  a  fraction  by  an  integer. 

Rule.  1st.  Cancel  all  factors  common  to  the  integer  and  the 
denominator. 


MULTIPLICATION  OF  Fli ACTIONS  87 

2d.  Multiply  the  numerator  by  the  remaining  factors  of  the 
integer. 

This  follows  from  Arts.  127  and  137. 

139.    Prob.      To  multiply  a  fraction  by  a  fraction. 

Rule.  1st.  Cancel  all  factors  common  to  either  numerator  and 
either  denominator. 

2d.  Take  the  product  of  the  remaining  factors  of  the  numerators 
for  the  numerator  of  the  result,  and  the  product  of  the  remaining 
factors  of  the  denominators  for  the  denominator  of  the  residt. 

Dem.     Let  it  be  required  to  multiply  -  by  — .  these  fractions 

b         n 

being  supposed  to  be  simple,  since  if  the  given  fractions  are  not 
simple,  they  may  be  made  so  without  changing  their  values. 

Multiplying  by  m  gives  (Art.  137)  ^.  But  we  were  to  mul- 
tiply, not  by  m,  but  by  one  nth.  of  m;  hence  this  product  must 
be  divided  by  n,  which  gives  (Art.  137)  — . 

The  rejection  of  all  factors  common  to  the  numerator  and 
denominator  will  evidently  leave  the  result  the  same  as  it  would 
have  been  had  they  been  rejected  before  the  multiplication. 
Their  rejection  before  multiplying  leaves  simpler  quantities  to 
multiply. 

EXAMPLES  XXXIX 

Multiply  the  following : 

±    x2-j-2x-S  b     4s2-12a-40 

x2  +  2  x  -  3     4  x2  -  12  x  -  40 

Operation.  — — x  — 

x*  +  5x  +  G      3x2-18z+15 

=  (x  +  S)(x  -l)y4(x-  5)(x  +  2)=  4 
(s  +  3)(x  +  2)     3(x-5)(x-l)     3 

2.   §££-  by  21a^.  3.    -v^— ,  by  x-y. 

28  xKyl  x-  —  yl 

^  +  4 5    _3  by  ^  +  2z-3. 

^  +  4a;-21     J  ^._7#  +  6     J 

Sue  Factor  the  denominator  of  the  5th  by  the  method  of  Art.  101,  first 
supplying  the  missing  term,  Ox2. 


88  HIGHER  ALGEBRA 


3o^     G^c3      10  ac2  ?         a        ,       a2  +  63 

4  c4        5  a6        9  63  a4-6«     ^   a2  -  W 

_     9  a,*2  —  1    ,      sb2  +  5  sb  _      a-  —  \     *         afi  —  1 

8.   by   — ! 9.  by  

xi-2ox     J    3sb-1  (a  +  1)2     J    (a2  -  a)2 

;v    &  +  3x-iB  .      2sb3-4sb2 
10.   — ! by 

a?  _  8  sb  +  12     J       af  -  6 

__    5x  +  W  ^  3x-9^  8x2-2 


8  a?  -  4       10  sb  +  5      3  sb2  -  27 


,  _    1  —  it*2      1  —  w2 

12. X  —n  X  [  1  + 


1  +  ?/       x  +  x2 


13.     x2~a;-6     by   _£-2»-8 


sb2  +  4sb  +  4         a;2 -7  sb +  12 

14     x2  +  ?/2   b     xy  -  y2 
x2  —  xy  x4  —  if 


"     +:-+ Uy 


iB"  —  ?/- 

2\x  —  y     %  -\-  yj  ~J   v?y  +  sb?/2 

16  rf  +  Xy-2tf    b     or  -  7  sb//  +  12  ?/2 
sb2  —  5  #2/  +  4  y2  x2  +  5  xy  +  6y2 

17  sb2  y2       ,       (sb2  —  ?/2)2 

"     SB2-/        SB2  +  ?/2       y    (f^ff+XJ+tff 

18  sb3  +  4sb2  +  4sb  +  3  ,       ar3  -  5  sb2  +  5  sb  +  2    . 
'    sb3  -  4  sb2 +  2  sb +  1      y       Xi-X2-X-2 

19  3  x4  -  17  sr3  +  27  sb2  -  7  sb  -  6    b     2  sr3  +  7  sb2  +  7  sb  +  2 

'   2  sb4 +  13  sr3  + 28  sb2 +  23  sb +  6     J    x?  -  6  sb2  +  11  x  -  6  " 

20  sb5  -  3  SB4  -  5  Xs  +  15  SB2  +  4  SB  -  12   . 

sb4  -  sb3  -  19  sb2  +  49  .b- 30      "     y 

aj«  -  2  sb3  -  13  sb2  +  38  sb  -  24 


a*  _  2  sb4  -  10  ar3  +  20  sb2  +  9  sb  -  18 


21    10s*ri-15sB3-14sB2  +  21   b     Bar8  -  9sb4  +  5sb  -  15 
15sb7-21sb4  +  25st3-35     y    2xi  -  6sb2-  3sb  +  9  ' 


DIVISION  OF  FRACTIONS  89 

DIVISION   OF   FRACTIONS 

140.  Prob.      To  divide  a  fraction  by  an  integer. 

Rule.  1st.  Cancel  all  factors  common  to  the  integer  and  the 
numerator. 

2d.  Multiply  the  denominator  by  the  remaining  factors  of  the 
integer. 

This  follows  from  Arts.  127  and  137. 

141.  Prob.     To  divide  a  fraction  by  a  fraction. 

Rule.     Invert  the  divisor  and  proceed  as  in  multiplication. 

Dem.     Let  it  be  required  to  divide  -   by  — ,   these  fractions 

b  n 

being  supposed  to  be  simple,  since  if  the  given  fractions  are  not 
simple,  they  may  be  made  so  without  changing  their  values. 
Dividing  by  m  gives  (Art.  137)  — .     But  we  were  to  divide, 

not  by  m,  but  by  one  nth.  of  m ;  hence,  since  the  divisor  used  is  n 
times  what  it  should  be,  the  quotient  is  one  nth  of  what  it  should 

be,  and  must  be  multiplied  by  n,  which  gives  (Art.  137)  — .    We 

observe  that  in  the  operation  we  have  inverted  the  divisor  and 
proceeded  as  in  multiplying  a  fraction  by  a  fraction. 

EXAMPLES   XL 

Divide  the  following : 

1.   ^-^  by  Ua2b2.  2.   <*-9f  by  ar>  +  3y. 

6xy-  x  +  ±y 

3    a  -.V  by  #  +  xy  +  y2  4>   « °_  by  a"  -  bm. 

3  ax  ab 

5    t^ll  by     a~b    •  6    *2  +  10*  +  21  by  £=£ 

'   a  +  2b    J  3a  +  6b  '    ^  -  4  or*  +  3  a    J  x*  -  x2 

7.   *~y*  by  *=!'•  8.   x--±  by  (a  +  xy\ 

tf    _|_    yS  *>  rg    _|_    y  Q  X  aX 

s»+4gy+4yg  b    xy  +  2y2  4  (a?  -  ah)  ,         C>ab 

x-y  y    tf-xy  '    b(a  +  bf      J  d2-b2 


90  HIGHER  ALGEBRA 


11.   a2  -  b2  -  c2  -  2  be  by  ^L±Jl±l. 
J   a+b-c 

2sf±lSx±W  ,      2  a;2 +  11  a; +  5 
4a;2 -9  y         4a;2- 1 

13    ar*  -  14  x  -  15  b     ar9- 12  a; -45 
x2  -  4  a;  -  45    1J   x2  -  6  x  -  27 ' 

14.  5  - a*  -  19 ^  by  3  -  a~5x. 

15.  ^  +  2.V  |  x  by  x  +  ty        ® 

x  +  y       y  y  x  +  y 

16.  f  ~4   by 


16         ar3  —  a;2  -f  4  a;  —  4 
17  a? -7  a; +  6  a^  +  3^ 


a!4-3s8-7s2  +  27a;-18     *      a;4  -  10  a;2  +  9 
18    2a;4  +  3ar3-14ar*-9a;  +  18  b    2a;4  -  9ar»  +  4ar*  +  21  x  -  18 
aJ4  +  a?-7aj2-13*-6        J    a;4  +  6 ar3  +  13a;2  +  12 x  +  4  ' 


SIMPLIFICATION   OF   COMPLEX   FRACTIONS 

142.  Prob.     To  reduce  a  complex  to  a  simple  fraction. 

Rule.  Multiply  numerator  and  denominator  of  the  complex 
fraction  by  the  1.  c.  m.  of  the  denominators  of  the  partial  fractions. 

Dem.  This  does  not  change  the  value  of  the  complex  fraction 
(Art.  127),  and  it  removes  the  partial  denominators,  since  they 
are  factors  of  the  quantity  by  which  all  the  partial  fractions  are 
multiplied. 

143.  Note.  A  complex  fraction  may  be  simplified  also  by 
uniting  the  terms  of  the  numerator  and  denominator  separately, 
and  then  dividing  the  former  result  by  the  latter. 

EXAMPLES  XLI 

Reduce  the  following  to  simple  fractions : 

a_  c  a      b 

b      d  b      a 


1. 


f*h  ab 


SIMPLIFICATION  OF  FRACTIONS  91 


a  +  h  +  'l  <i±A  +  >±=± 
a  a  —  1      «.  +  1 

6  a  —  1      a  +  1 


^    x  +  y         x                                                     a  +  4 
*•    ; — •  6.    — — • 

a  +  1  - 


x—yx  a + 4 

a  ,    b  ,    c 


m  —  n  .    m* 


+  _  +  I^__L"  + 

6c     ac     ao  m  +  « 


7. -•  8. 


aft     ac  ,  6c  m2        nt*»  -f  n3 

c        b       a  m  —  n      (ra  —  n)2 

144.    Prob.      To  simplify  a  terminated  continued  fraction. 
Rule.    Begin  at  the  bottom  and  simplify  backwards  and  upwards, 
performing,  step  by  step,  the  operations  indicated. 

EXAMPLES   XLII 
Simplify  the  following : 

1  a 


l. 


*  +  -~T  b-r-—^- 


1 


y+-  d+ 

3.    


e 


i 


x  +  2 


a'  — 


.+*-£±i  a+ 


*+     >' 


a  +  1 


1  +  7 

5.    T^^-b"). 


b 
a x  + 


1_a  1+*±I 

&  3 -a 

145.    Prob.     To  free  a  fraction  from  negative  exponents. 

Rule.     If  the  quantity  affected  with  a  negative  exponent  is  a 
factor  of  the  numerator,  transfer  it  to  the  denominator,  or  if  a 


92  HIGHER  ALGEBRA 

factor  of  the  denominator,  transfer  it  to  the  numerator,  and  change 
the  sign  of  the  exponent. 

If  it  is  a  term  or  part  of  a  term  of  a  polynomial,  transfer  it  to 
the  denominator  of  the  term  in  which  it  stands,  changing  the  sign  of 
the  exponent,  and  then  reducing  the  resulting  complex  fraction  to  a 
simple  one;  or,  what  is  the  same  thing,  without  this  transference 
multiply  both  numerator  and  denominator  by  the  same  quantity 
affected  ivith  a  numerically  equal  positive  exponent. 

Dem.  The  rule  is  a  direct  consequence  of  the  signification  of 
a  negative  exponent  (Art.  9),  viz.  the  reciprocal  of  what  the  ex- 
pression would  be  if  the  exponent  were  positive. 

EXAMPLES  XLIII 
Free  the  following  from  negative  exponents : 

±    a-2b-3c\ 
x~xy~n ' 

3.  (x~y)~\ 

x~3  +  y~s 

(ar1  -  y~2)~2  (x~2-iy 


2 

a~' 

l-b- 

-3 

a~{ 

5+6- 

-2 

4 

xy 

-3(a~ 

2-b 

-2) 

z~ 

2  (a"3 

■  +  b~ 

3) 

6. 

x-'' 

m(x2- 

-</)" 

■n 

CHAPTER  IX 

THEORY  OF  EXPONENTS,   INVOLUTION,   AND  EVOLUTION 

SECTION   I  — THEORY   OF  EXPONENTS 

146.  We  may  either  define  fractional  and  negative  exponents, 
as  in  Art.  9,  and  prove  the  index  law  of  multiplication  to  be 
general,  as  in  Art.  48,  or  we  may  assume  the  index  law  as  proved 
for  positive  integral  exponents  to  be  general,  and  determine  what 
fractional  and  negative  exponents  should  mean. 

147.  Theorem.  If  the  index  law  of  multiplication  is  made 
general, 

1st.  The  numerator  of  a  fractional  exponent  denotes  a  power, 
and  the  denominator  a  root. 

2d.  A  negative  exponent  denotes  the  reciprocal  of  ivhat  it  would 
denote  if  positive. 

Dem.     1st.    By  definition  (Art.  9,  2d  (a)), 


(icn)n  =  X"  •  xn  •  of  •••  to  n  factors.  (1) 

By  the  index  law  of  multiplication  (Art.  48,  1st)  this  is 

—  H 1 — ton  terms  ,_, 

xn    n    n  =  xm.  (2) 

m 

Therefore,  (x7l)n  =  xm.  (3) 

Extracting  the  nth  root  of  these  equals,  we^have 

m 

x"  =  y/gr.  (4) 

2d.    By  the  index  law  of  multiplication  (Art.  48), 

x11  •  x~n  =  t?  s  1  (Art.  70). 
93 


94  HIGHER  ALGEBRA 

Dividing  these  equals  by  xn,  we  have 

x~n  =  — • 
xn 

EXAMPLES  XLIV 

Express  the   following   without   fractional   and   negative   ex- 
ponents : 

1.   a¥.  2.   aW.  3.   5(ax)~*. 

i 

2  _    5x~$  -    7*x  " 

O.     •  D. 


7. 


3  x*  7  2/*  6*  y? 

m 


3a26"2  Q    5~2^Vf 

■ •  O.         ■ ' *  w. 

5*  a"*6*  fSfi  4-i  yn 


Express  the  following  without  radical  signs  and  negative  ex- 
ponents : 

io.  v^^?.  ii.  tv&y?.      i2.  vzw^r*. 

JLo.      — — _  •  Xrk,     — —     _ ^^z —  * 

Vsc  v^2/"2  V5  Va  V#-3  V?/2 

15.    m^gy^L-  16.    ^/(^T67V(a-6)-3. 

tyx  t  Vy~* 


1?_    ^(a2  +  2  5)\  18_    V(^-y2)-^ 

V(x  +  i/)-3  *    ^/(^  +  2/2)^ 

148.  Theorem.  The  product  of  several  quantities  affected  collec- 
tively by  an  exponent  is  equal  to  the  product  of  the  several  quantities 
affected  separately  by  the  same  exponents. 

Dem.     1st.   When  the  exponent  is  a  positive  integer. 
(abc~-)n 

=  {abc  '")(abc  >-)(abc  ...)...  to  n  factors, 

=  (aaa  •••  to  n  f actors) (bbb  •••  to  n  f actors) (ccc  •••  to  n  factors)  •  ••> 

=  anbncn  •••. 


THEORY  OF  EXPONENTS  95 

2d.   When  the  exponent  is  a  positive  fraction. 

m    to  m  to  to  to 

(a^&V  ...)*  =(a-)"(6*)"(c*)"  •••  by  the  1st  case, 

=  a^c"1  —by  (3)  of  Art.  147, 

=  {abc  •••)"*  by  the  1st  case. 

Extracting  the  nth  root  of  these  equals,  we  have 

TO      TO     TO  TO 

anbncn  ~>  =  (abc  •••)•. 

3d.   When  the  exponent  is  negative. 

By  Art.  71,  and  the  1st  case  of  this  demonstration, 

(abc  ..•)""  = = =  a  nb-nc-n  .... 

v  J         (abc  •••)"     anbncn  ••• 

149.   Cor.     By  the  proposition, 

fa\n     (       IV       „      1      an 

[-)  ±    ox7     =anx—  =  —  ; 

\6y     V     &y  hn    bn' 


also,  \h  =  \a     h~*a 


la      n\     ~i      „/-  v^ 


150.  Sch.  The  above  theorem  and  corollary  are  briefly  stated 
thus :  The  power  of  the  product  equals  the  product  of  the  powers, 
and  the  root  of  the  product  equals  the  product  of  the  roots  (the 
indices  being  the  same)  ;  also  the  power  of  the  quotient  equals  the 
quotient  of  the  poivers,  and  the  root  of  the  quotient  equals  the  quotient 
of  the  roots  (the  indices  being  the  same). 

The  student  needs  to  be  cautioned  against  supposing  that  the 
power  or  root  of  the  sum  or  difference  equals  the  sum  or  differ- 
ence of  the  powers  or  roots. 

151.  Prob.      To  affect  a  monomial  with  any  exponent. 

Rule.  Multiply  the  exponent  of  each  of  the  factors  by  the  given 
exponent. 

Dem.  Let  it  be  required  to  affect  af*  with  the  exponent  n,  m 
having  any  value,  integral  or  fractional,  positive  or  negative. 


96  HIGHER  ALGEBRA 

1st.    When  n  is  a  positive  integer. 

(xm)n  =  x^xmxm  •••  to  n  factors, 

=  xn 

riffnn 


/y.w+m+mH —  to  n  terms 


2d.   When  n  is  a  positive  fraction,  as  —  • 

p 
By  Art.  147,  (xm)q  is  the  gth  root  of  the  pth.  power  of  xm,  i.e., 


(xny  =V(xmy, 

_  -fyo/mp^  \yj  the  1st  case  of  this  demonstration, 

mp 

=  a?« ,  by  Art.  147. 

3d.   When  n  is  negative. 

By  Art.  147,  and  the  1st  case  of  this  demonstration, 

(xm)~n  =  -i—  =  —  =  x~mn. 

Finally,  by  Art.  148,  what  has  been  proved  above  for  xm  applies 
to  any  number  of  factors  of  any  monomial. 

EXAMPLES   XLV 
Perform  the  following  indicated  operations : 
l.   (a362)2.  2.   (a-V2c*)2.  3.   (ah~k3f. 

4.    (32afy-*)3.  „      5.    (5x^srl)\  6.    (6*«yW)5. 


WJ  \VyV  \  ®>Vi J 

10.    (oafy6)*.  11.    (25a4#V2)*.  12.    ' 


\27  by*, 
13.   (5^-y)"3.  14.    (a~%-y)"i  15.    (32  a5b-l0c^)\ 


16.    (-64ic62/Vw)i  17.  {V(afV)8}  *•        18    (^m92/^7?- 


BINOMIAL    THEOREM  97 


SECTION  II  — INVOLUTION 


152.  Involution  is  the  process  of  raising  quantities  to  required 
powers. 

153.  Prob.     To  raise  a  monomial  to  any  required  power. 

Rule.  Use  the  numerical  coefficient  as  many  times  as  a  factor 
as  there  are  units  in  the  degree  of  the  power,  and  multiply  the  expo- 
nent of  each  literal  factor  by  the  exponent  of  the  required  power, 
writing  the  minus  sign  before  the  result  only  when  an  odd  power  of 
a  negative  quantity  is  required. 

This  is  but  an  application  of  the  definition  of  a  power  (Art.  9), 
affecting  a  monomial  with  any  exponent  (Art.  151),  and  the  law 
of  signs  in  multiplication  (Art.  47). 

154.  Factorial  n  is  the  product  of  all  the  integral  numbers  from 
1  to  n  inclusive,  and  is  written  \n.* 

Thus,  [3  =  2-3,   [5=2.3-4.5,  |w  =  2  •  3  •  4  ...  n. 

155.  Prob.      To  raise  a  binomial  to  any  required  power. 

Formula,  x,  y,  and  m  being  any  quantities  whatever,  positive 
or  negative,  integral  or  fractional, 

(x+y)*»  =  x>»+mx<»->y+  m(m~1)^-y+  Mm-l)(m-2)xtn_y 
+  m(m-l)(m-2)(m-3)  ^_y  +  ^ 

This  is  Newton's  Binomial  Formula,  the  demonstration  of  which 
is  given  in  a  subsequent  part  of  the  work.  From  an  inspection 
of  the  formula  we  deduce  the  following  theorem,  called 

THE   BINOMIAL   THEOREM 

In  the  expansion  of  a  binomial  affected  with  any  exponent,  the 
exponent  of  the  first  letter  begins  in  the  first  term  with  the  exponent 


*  The  notation  n  !  is  also  employed. 
downey's  alg.  —  7 


98  TITO  HER  ALGEBRA 

of  the  binomial,  and  in  each  succeeding  term  decreases  by  1 ;  while 
the  exponent  of  the  second  letter  begins  in  the  second  term  with  1, 
and  in  each  succeeding  term  increases  by  1. 

Tlie  coefficient  of  the  first  term  is  1 ;  that  of  the  second,  term  is  the 
exponent  of  the  binomial;  and  if  the  coefficient  of  any  term  be  mul- 
tiplied by  the  exponent  of  the  first  letter  in  that  term  and  divided  by 
the  exponent  of  the  second  letter  increased  by  1,  the  result  will  be  the 
coefficient  of  the  next  term. 

156.  Cor.  i.  The  expansion  of  a  binomial  terminates  when  m 
is  a  positive  integer,  and  the  number  of  terms  is  m  -j-  1. 

For  the  coefficient  m(m  —  l)(m  —  2)  •••  (m  —  m),  which  is  in 
the  (m  +  2)th  term,  and  all  subsequent  coefficients,  are  0. 

157.  Cor.  2.  When  m  is  a  positive  integer,  the  coefficients  equally 
distant  from  the  extremes  are  numerically  equal. 

For  the  expansion  of  (b  +  a)m  has  the  same  value  as  the  expan- 
sion of  (a  +  b)m,  but  the  terms  occur  in  the  reverse  order.  Hence 
beyond  the  middle,  which  is  a  term  or  a  sign,  according  as  m  is 
even  or  odd,  the  coefficients  need  not  be  computed. 

158.  Cor.  3.  If  the  sign  of  the  second  letter  is  minus,  and  m  is 
a  positive  integer,  the  signs  of  the  terms  of  the  expansion  will  be 
alternately  plus  and  minus. 

For  the  odd  powers  of  a  minus  quantity  are  minus. 

159.  Note.  At  this  stage  of  his  progress  the  student  should 
learn  the  above  theorem  and  familiarize  himself  with  its  use  in 
raising  binomials  to  powers,  i.e.  for  positive,  integral  exponents. 
He  should  be  careful  to  note  that,  while  such  a  form  as  (3  a2— 2  bs)5 
may  be  expanded  by  the  theorem,  the  laws  apply,  not  to  the  expo- 
nents and  coefficients  of  a  and  b  in  the  full  expansion,  but  to 
those  of  3  a2  and  —  2  bs,  regarded  as  the  terms  of  the  binomial. 
These  should  be  kept  in  parentheses  through  the  expansion,  the 
indicated  operations  being  performed  afterward.  It  must  not  be 
forgotten  that  an  odd  power  of  a  negative  quantity  is  plus  and 
an  even  power  minus. 


J  « 

J. 

-/  /ft- 

f 

<-&-' 

BINOMIAL   THEOREM 


EXAMPLES  XL VI 
Expand  the  following  by  the  Binomial  Theorem : 

1.  (3x2-2tfy\  v__ _ ___~ 

Solution.  Here  3  x2  takes  the  place  of  the  first  letter,  and  '  —  2  yz  of  the 
second  letter  of  the  formula.  ~*~c*L&  r 

Hence  we  have  by  the  theorem, 

(3  x2  -2  ifY  =  (8  x°')5  +  &(3  x2)4(-  2  ?/3)  +  10(3  x2)3(-  2  y*)2       . 

+  10(3  X2)2(-  2  *»)«  +  5(3  x2)(_  2  y3)4  +  (_  2  y3)5 

=  243  x10  -  810  xV  +  1080  x6^  -  270  x*y»  +  240  x2y12  -  32  y15 

Since  wt  =  5,  the  expansion  will  contain  6  terms  (Art.  156),  and  the  coeffi- 
cients after  the  middle  sign  are  the  same  as  those  before  it,  but  in  the  reverse 
order,  and  may  be  written  without  further  computation  (Art.  157). 

2.  (x-\-y)\  3.  (a -b)5.  4.  (2x*-yy. 

5.  (l-x2)4.  6.  (a  +  by.  7.  (x*-y2y. 

8.  (x*  +  2y)4.  9.  (l+2ar9)5.  10.  (a*  +  2  fc^"1)4. 

11.  (.*;-i)6.  12.  (-J/a+V6)8.  13.  (ix-2-2aV)6. 


14.   (.k  -  g  +  z)4.      15-   (V^  +  l-Va-l)4.      16.   (x  +  V^-l)6. 

160.  A  special  theorem  for  squaring  a  polynomial  has  been 
given  in  Art.  64.  It  is  convenient  to  have  also  a  special  theorem 
for  cubing  a  polynomial. 

161.  Theorem.  The  cube  of  any  polynomial  is  the  algebraic  sum 
of  the  cubes  of  all  the  terms,  three  times  the  square  of  each  term  into 
each  of  the  other  terms,  and  six  times  each  group  that  can  be  formed 
icith  three  terms  each  as  factors. 

Dem.  When  there  are  more  than  two  terms,  part  of  them  may 
be  treated  as  constituting  the  first  term,  and  the  others  as  the 
second  term  of  a  binomial,  thus : 

(a  +  b  +  cy=[(a  +  b)+cy=(a+by+3(a  +  b)2c+3(a  +  by+c*      . 

=  a3-f 3  d2b+3  a62+6s+ 3  a2c+(5  abc+3  b2c+3  ac2+3  bc2+(? 
=as+bs+c*+3cfb+3a2c+3b2a+3b2c+3c2a+3c2b  +  6abc. 


100  HIGHER   ALGEBRA 

Again,  (a+b  +  c—<Xf=[(a  +  b)  +  {c—d)J 

=  (a-|-6)8+3(a+6)2(c-d)  +  3(a+6)(c-d)2+(c-d)8 

=a3+63+c3-d3+3a26+3a2c-3a2d  +  3&2a+3  62c 

-3  tfd+3  c2a+3  c26-3  c2d+3  d2a+3  d2b+3  d2c 

+  6  abc—6  abd—6acd—6bcd. 

By  inspecting  the  results  of  these  operations  we  deduce  the 
important  theorem  stated  above. 

EXAMPLES   XLVII 

Cube  each  of  the  following : 
1.  4£2  +  2#-3. 

Operation.     It  is  best  first  to  write  in  order  the  different  powers  of  x, 
and  then  supply  the  coefficients  in  the  order  of  the  rule,  thus  : 

x6  x5  x4  x3  x2  x 

64  96    -144  8      -  36  54     -  27 

48      -  144         108 


64  x6  +  96  x5  -  96  x4  -  136  x3  +  72  x2  +  54  x  -  27 

2.   2x2-3x-\-5.  3.   6a?-4xy  +  2y2. 

4.  4a?-2<B2  +  3»-f-5.  5.   x4-2x*  +  3x2-x  +  2. 

SECTION  III  — EVOLUTION 

162.  Evolution  is  the  process  of  extracting  roots  of  quantities. 
In  extracting  any  even  root  of  a  quantity  there  are  always  two 

results,  differing  only  in  sign,  since  even  powers  of  both  positive 
and  negative  quantities  are  plus.  For  the  present  purposes  it  is 
customary  to  write  only  the  positive  root,  or  in  case  of  polyno- 
mials, the  root  that  begins  with  a  positive  term. 

163.  Prob.      To  extract  any  root  of  a  monomial. 

Rule.     Extract  the  required  root  of  the  coefficient  and  divide  the 
exponent  of  each  letter  by  the  index  of  the  root. 


SQUARE  ROOT  OF  POLYNOMIALS  101 

Dem.  This  is  but  an  application  of  Art.  151,  since  to  extract 
the  square  root  is  to  affect  a  quantity  with  the  exponent  \,  the 

cube  root  -J,  the  nth  root  -,  etc. 

164.  Prob.     To  extract  a  root  whose  index  is  a  composite  number. 

Rule.  Extract  successively  the  roots  indicated  by  the  prime  fac- 
tors of  the  index. 

Dem.     By  Arts.  147  and  151, 

2_  l  i  - 

m  mVx  =  xmn  =  (af)»  =  V^a>; 

j_  l  i    .. 

also  mVx  =  xmn  =  (xn)m  =  V  Vaj. 

Hence  the  4th  root  is  the  square  root  of  the  square  root ;  the 
6th  root  is  the  cube  root  of  the  square  root,  or  the  square  root  of 
the  cube  root ;  the  8th  root  is  the  square  root  of  the  square  root 
of  the  square  root ;  the  9th  root  is  the  cube  root  of  the  cube  root ; 
and  so  on. 

SQUARE  ROOT  OF  POLYNOMIALS 

Case  I 

165.  When  the  square  is  a  trinomial. 

Rule.  Arrange  with  reference  to  one  letter,  extract  the  square 
root  of  the  extreme  terms,  and  connect  the  results  by  the  sign  of  the 
middle  term. 

This  follows  from  Arts.  61  and  62.  The  expression  is  not  a 
perfect  square  unless  the  middle  term  is  twice  the  product  of  the 
square  roots  of  the  extreme  terms. 

Case  II 

166.  When  the  square  contains  only  tivo  powers  of  any  one  of  its 
letters. 

Rule.  Arrange  ivith  reference  to  any  letter  which  has  only  two 
poioer*.  regard  the  terms  not  containing  that  letter  as  the  third  term 
of  a  trinomial,  and  proceed  as  in  the  preceding  case. 


102  HIGHER  ALGEBRA 

EXAMPLES  XLVI1I 
Find  the  square  root  of  each  of  the  following : 

1.  9  a4  +  4  b2  +  c6  -f  12  a2b  -  6  a2c3  -  4  6c?. 

Solution.  Here  we  have  but  two  powers  of  each  letter,  viz.,  a4  and  a2, 
b'2  and  &,  c6  and  c3.  Arranging  with  reference  to  a  (any  other  letter  would 
do  as  well),  we  have 

9  a4  +  (12  b  -  6  c3)a2  +  (4  b2  -  4  6c3  +  c6). 

The  square  root  of  the  first  term  is  3  a2,  and  of  the  last  term  2  b  —  c3 ; 
while  the  middle  term  is  plus  twice  the  product  of  these  square  roots.  Hence 
the  square  root  is  3  a2  +  2  b  —  c3. 

2.  a2  +  & W  -  2  abVd3  +  bYd*  +  2  abed  -  2  6W. 

Solution.  Here  a  is  the  only  letter  having  but  two  powers.  Arranging 
with  reference  to  this,  we  have 

a2  +  (2  bed  -  2  6W8)a  +  (&2c2d2  -  2  6W*  +  &4c4d3) 

or  a2  +  (2  bed  -  2  &2c2#)a  +  (bed  -  MAP)*. 

Hence  the  square  root  is  a  +  6ccZ  —  &&&. 

3.  4  a2  +  9  &4  -  12  aft2  -  24  &2d3  +  16  ads  +  16  d\ 

4.  16  sc4  4- 16  oH  4-  ?/2  —  8  ary  4-  4  z2  —  4  ys. 

5.  25z4  +  20^-4#z2  +  4/  +  z4-10a;2z2. 

6.  9x^z2  +  12xY-4:tfz-\-4y4-Gxiz. 

7.  16  a4  4-  24  a2b  +  9  b2  -  16  a2c3  4-  4  c6  +  8  a2d  4-  dJ  -  12  &c3  + 
6  6d-4c3d. 

Solution.  Here  each  letter  has  but  two  powers.  Arranging  with  refer- 
ence to  a  and,  at  the  same  time,  arranging  with  reference  to  b  the  terms  that 
do  not  contain  a,  we  have 

16  a*  +  (24  b  -  16  c3  +  8  d)a2  +  [9  V2  +  (6  d  -  12  c3)&  +  (4  c6  -  4  c3d  +  <P)]. 

The  square  root  of  the  first  term  being  4  a2  and  of  the  last  term  (the  part 
in  the  brackets)  3  b  —  2  c3  +  d,  and  the  middle  term  plus  twice  the  product 
of  these  square  roots,  the  square  root  of  the  whole  expression  is 

4  a2  +  3  b  -  2  c3  +  d. 

8.  9  a6  -  6  a3&2  -  12  a3c  +  24  a*d  4-  64  4-  4  &*c  -  8  &2cJ  4-  4  c2  -  16  cd 
4- 16  d2. 


SQUARE  ROOT  OF  POLYNOMIALS  103 

9.  36  aB  +  30  «3c  -  24  aV  -  60  crtF  +  4  64  +  20  62d4  - 12  b2c  +  9  c2 
-30ctf4  +  25d8. 

10.  125  «9  +  225  aftr  -  150  a«c  +  135  <W  -  180  a362c  +  60  a3c2  + 
27  66  -  54  64c  +  36  W  -  8  c3. 

Case  III 
167.    Any  perfect  square  not  included  in  Cases  I  and  II. 
Solution.     By  Art.  64  we  have,  after  arranging  in  descending 
powers  of  x, 

(axn  +  bar-1  +  cxn~2  +  dxn~3  -\ )2 

=  a*s*  +  2  abx2'1-1  +  (b2  +  2  ae)  x2n~2  +  (2  6c  +  2  ad) x2^3  -f- .... 

In  squaring  a  polynomial  of  the  nth  degree  a  term  in  arn  can 
result  only  from  the  square  of  the  first  term  of  the  root;  hence  if 
we  take  the  square  root  of  the  first  term  of  the  arranged  square,  we 
shall  have  the  first  term  of  the  square  root. 

A  term  in  x2""1  can  result  only  from  twice  the  first  term  of  the 
root  by  the  second ;  hence  if  we  divide  the  second  term  of  the  ar- 
ranged square  by  tvnce  the  first  term  of  the  root,  we  shall  obtain  the 
second  term  of  the  root. 

Terms  in  x2'1-2  can  result  only  from  the  square  of  the  second 
term  of  the  root  and  twice  the  first  by  the  third ;  hence  if  we  sub- 
tract from  the  third  term  of  the  arranged  square  the  square  of  the 
second  term  of  the  root  and  divide  the  remainder  by  twice  the  first 
term  of  the  root,  we  shall  obtain  the  third  term  of  the  root. 

Terms  in  x2"'3  can  result  only  from  twice  the  second  term  of 
the  root  by  the  third  and  twice  the  first  by  the  fourth ;  hence  if 
we  subtract  from  the  fourth  term  of  the  arranged  square  twice  the 
second  term  of  the  root  by  the  third  and  divide  the  remainder  by 
twice  the  first  term  of  the  root,  we  shall  obtain  the  fourth  term  of  the 
root. 

In  the  same  way  four  terms  may  be  obtained  from  the  other  end 
of  the  square,  i.e.,  obtained  as  they  would  be  if  the  square  were  writ- 
ten the  other  end  foremost,  the  signs  being  left  temporarily  ambiguous. 

In  practice  it  is  better  to  obtain  terms  alternately  from  the  two 
ends,  continuing  the  operation  only  until  reaching  d  term  that  is 
numerically  the  same,  from  whichever  end  obtained.     The  sign  of 


104  HIGH  Eli  ALGEBRA 

this  term  as  obtained  from  the  first  end  will  determine  the  signs  of 
the  terms  obtained  from  the  last  end. 

168.  Sch.  1.  If  the  square  of  the  second  term  of  the  root  is 
similar  to  the  fourth  term  of  the  given  polynomial  instead  of  the 
third,  the  third  term  of  the  root  is  found  by  dividing  the  whole 
of  the  third  term  of  the  given  polynomial  by  twice  the  first  term 
of  the  root. 

If  the  square  of  the  second  term  of  the  root  is  similar  to  a 
missing  term  of  the  given  polynomial  between  the  second  and 
third  terms,  the  third  term  of  the  root  is  found  by  subtracting 
the  square  of  the  second  term  of  the  root  from  0,  the  coefficient 
of  the  missing  term,  and  dividing  the  remainder  by  twice  the  first 
term  of  the  root. 

169.  Sch.  2.  Test.  If  the  overlapping  term  (i.e.,  the  term  con- 
taining the  same  power  of  x  from  whichever  end  obtained)  is  not 
numerically  the  same  when  found  from  each  end,  the  polynomial 
is  not  a  perfect  square.  When  it  is  the  same,  to  ascertain  whether 
the  polynomial  is  a  perfect  square,  it  is  necessary  simply  to  note 
how  those  terms  not  used  in  the  operation  are  made  up  from  the 
square  of  the  supposed  square  root. 

170.  Note.  The  advantage  of  this  method  over  that  employed 
for  extracting  the  square  root  of  numbers  is  its  brevity,  the  square 
root  being  written  at  once,  without  writing  any  intermediate 
steps.  The  written  work  in  the  model  solutions  below  is  merely 
for  explanation  and  can  all  be  omitted  in  practice. 

EXAMPLES  XLIX 
Find  the  square  root  of  each  of  the  following: 
1.   25a4-30ar3  +  49a2-24a  +  ±6. 
Solution.     If  this  is  a  perfect  square, 

first  term  of  sq.  rt.  =  V25&4  =  5  x2, 
second  term  of  sq.  rt.  =  —  30  x3  -=-2(5  x2)  =  —3  se, 
last  term  of  sq.  rt.  =  VlO  =  ±  4, 
next  to  last  term  of  sq.  rt.  =  -  24  x  -=-  2  (  ±  4)  =  T  3  x. 


SQUARE  ROOT  OF  POLYNOMIALS  105 

This  term  (=F  3x)  is  numerically  the  same  as  the  second,  and  to  make  it 
the  same  in  sign  the  upper  signs  must  be  used.  Hence  the"  square  root  is 
5  X2  _  3  x  +  4. 

Test.  In  the  operation  all  the  terms  of  the  given  polynomial  have  been 
used  except  49  x2.  By  reference  to  Art.  64  it  will  be  seen  that  in  squaring 
5  x2  —  3  x  +  4,  terms  in  x'2  can  result  only  from  the  square  of  the  second  term 
and  twice  the  first  by  the  third.  This  gives  (-  3x)2  +  2(5  x)  (4)  =  49  x2, 
showing  the  given  polynomial  to  be  a  perfect  square. 

2.  49a;4 +  28^ -38a,*2 -12a; +  9. 

3.  36a-6  -  36a? -3a-4  + 6a,-3  +  a*2. 

4.  16  -40a;  +  89ar*-80ar}  +  64a* 

5.  8a;  +  4  +  a;4  +  4ar3  +  8a*2. 

6.  o;4-ar3  +  -+4a;-2+4- 

4  or 

7.  36«4-48ar>  +  52afy2-24#?/3  +  92A 
Solution.     If  this  is  a  perfect  square, 

first  term  of  sq.  rt.  =  V36x*  =  6  x2, 

second  term  of  sq.  rt.  =  —  48  xhj  +  2(6  x2)  =  —  4  xy, 

last  term  of  sq.  rt.  =  V9  y*  =  ±  3  y2, 

next  to  last  term  of  sq.  rt.  =  —  24  x?/3  -h  2(  ±  3  y2)  =  T  4  xy. 

We  see  that  there  are  but  three  terms  and  that  the  plus  sign  of  \A)y* 
must  be  used.     Hence  the  square  root  is  6  x2  —  4  xy  +  3  y2. 

Test.     (4  xy)2  +  2(6  x2)  (3  y2)  =  52  x2y2. 

8.  25a?4-30arfy-31a%2  +  24a^  +  16?/4. 

9.  9z4  +  24afy-14afy2-40a;<7-3  +  25?/4. 

10.  16a8  _32afy2-40 xy  +  56z?f  +  49/. 

11.  «8  +  2ar5  +  2a-4  +  arJ  +  2a;  +  l. 

12.  x«-§x5  +  9 as*  —  10 a;8  +  30 x*  +  25. 

13.  49  x™  -56x'  +  16  x°  +  126  av  -  72  a;3  4-  81. 

14.  9xr>  +  6a?*-  11  a;1  +  20  ar*  +  12  a2  -  16a;  4- 16. 


106  HIGHER  ALGEBRA 

Solution.     If  -this  is  a  perfect  square, 

first  term  of  sq.  rt.  =  V9  x6  —  3  x3, 
second  term  of  sq.  rt.  =  6  x5  -f-  2(3  x3)  =  x2, 
third  term  of  sq.  rt.  =  [  -  11  x4  -  (x2)2]  -  2(3  x3)  =  -  2  x, 
last  term  of  sq.  rt.  =  Vl6  =  ±  4, 
next  to  last  term  of  sq.  rt.  =  —  16  x  •*-  2(  ±  4)  =  T  2  x. 

This  term  (T2x)  is  numerically  the  same  as  the  third,  and  to- make  it 
the  same  in  sign  the  upper  signs  must  be  used.     Hence  the  square  root  is 

3x3  +  x2-2x  +  4. 

Test.  All  the  terms  of  the  given  polynomial  have  been  used  except  20  x3 
and  12  x2.  In  squaring  3  x3  +  x2  —  2  x  +  4,  terms  in  x3  can  result  only  from 
twice  the  first  by  the  fourth  and  twice  the  second  by  the  third,  giving 

2(3x2)  (4)  +  2(x2)(  -  2  x)  =  20  x3. 

Terms  in  x2  can  result  only  from  the  square  of  the  third  and  twice  the 
second  by  the  fourth,  giving 

(-2x)2  +  2(x2)(4)=12x2. 

Hence  the  given  polynomial  is  a  perfect  square. 

15.  9X6  -  24x5  +  28  x4  -46a?  +  44a2-  20x  +  25. 

16.  16x6-40x5  +  41x4-44x3  +  34x2-12x  +  9. 

17.  49  x6  +  42  x5  -  19  x4  -  68  x3  -  20  x2  +  16  x  +  16. 

18.  36  tf  -  48  afy  -  20  afy8  +  84  xhf  -  31  x2?/4  -  30  xy5  +  25  if. 

19.  64  xs  -  16  x5  -  47  x4  -  90  x3  +  21  x2  +  36  *  +  36. 

20.  121  x6  -  66  afy  +  119  xy  + 168  xY  -  29  x2?/4  +  90  x,?/5  +  81  tf. 

21.  9  x6  +  6  x>y  -  11  xSf  +  20  x3?/3  +  12  x2y*  -  16  xy5  + 16  /. 

22.  4x°  +  16x3-40x3  +  16x  +  4. 

Solution.     If  this  is  a  perfect  square, 

first  term  of  sq.  rt.  =  V4  x6  =  2  x3, 
second  term  of  sq.  rt.  =  16  x5  ■*-  2(2  x3)  =  4  x2, 
third  term  of  sq.  rt.  =  [0  -  (4  x2)2]  -*■  2(2  x3)  =  -  4  x, 
last  term  of  sq.  rt.  s=  VI  =  ±  2, 
next  to  last  term  of  sq.  rt.  =  16  x  ■+■  2(  ±  2)  =  ±  4  x. 


8QUABS   HOOT  OF  POLYNOMIALS  107 

This  term  (±4x)  is  numerically  the  same  as  the  third,  and  to  make  it 
the  same  in  sign  the  lower  signs  must  be  used.  Hence  the  square  root  is. 
2  x3  +  4  x2  -  4  x  -  2. 

Test.  All  the  terms  of  the  given  polynomial  have  been  used  except 
—  40  x3  and  a  missing  term  in  x2.  In  squaring  2  x3  +  4  x2  —  4  x  -  2,  terms 
in  x3  can  result  only  from  twice  the  first  by  the  fourth  and  twice  the  second 
by  the  third,  giving 

2(2  x3)  (  -  2)  +  2 (4  x2)  ( -  4  x)  =  -  40  x3. 

Terms  in  x2  can  result  only  from  the  square  of  the  third  term  and  twice 
the  second  by  the  fourth,  giving 

(_4x)2  +  2(4x2)(-2)=0. 
Hence  the  given  polynomial  is  a  perfect  square. 

23.  ars  +  2a6  +  2a~5  +  3a:4  +  2arJ  +  3arJ  +  2a;  +  l. 

24.  25^  +  20 a7  -  6 x6- 4. ^  +  61  a4 +  24. ^-12^ +  36. 

25.  16a,-10-  8 a?  +  16«7  +  17 x6-  28 ar5  + 14  ar5  -  8  ar*  -12 a?  +  9. 

Solution.     If  this  is  a  perfect  square, 

first  term  of  sq.  rt.  =  Vl6  x10  =  4  x5, 
second  term  of  sq.  rt.  =  —  8  x8  -*•  2(4  x5)  =  —  x8, 
third  term  of  sq.  rt.  =  16  x7  -5-  2(4  x5)  =  2  x2, 
last  term  of  sq.  rt.  =  V9  =  ±  3, 
next  to  last  term  of  sq.  rt.  =  —  12  x  ■*■  2(  ±  3)  =  T  2  x, 
third  from  last  term  of  sq.  rt.  =  [  -  8  x2  -  (  T  2  x)2]  +  2(  ±  3)  =  T  2  x2. 

This  term  (=F  2  x2)  is  numerically  the  same  as  the  third,  and  to  make  it 
the  same  in  sign  the  lower  signs  must  be  used.  Hence  the  square  root  is 
4  x5  -  x3  +  2  x2  +  2  x  -  3. 

As  the  square  of  the  second  term  of  the  root  is  similar,  not  to  the  third 
term  of  the  given  polynomial,  but  to  the  fourth,  the  whole  of  the  third  term 
must  be  divided  by  twice  the  first  term  of  the  root. 

26.  a;8  -  6  a;7  +  5  a*  +  22  u"  -  18  x*  -  44  x*  +  9  x2  +  40  x  +  16. 

27.  25x,0-30a^  +  20x7-31a^ 

+  48  x>  +  28  a-4  -  52  x*+  40  x*  -  48  x  +  36. 

28.  16  x™  -48  a;10 +  16  0*  +  12  a,-8 

-  24 x7  +  112 a:6 -  12 x5  -  99 x4  +  36 a?-  54 3?  +  81 

29.  16  a-10  -40  a*  +  41  a,-8  +  4.  x7 

-  74  a*  +  96  ar5  -  45  x4  -  24  a*  +  54  a?  -  36  x  +  9. 


108  HIGHER  ALGEBRA 

Solution.     If  this  is  a  perfect  square, 

first  term  of  sq.  rt.  =  VW  x10  =  4  x5, 
second  term  of  sq.  rt.  =  —  40  x9  -r-  2(4  x5)  =  —  5  x4, 

third  term  of  sq.  rt.  =  [41  x8  -  (  -  5  x4)2]  +  2(4  x5)  =  2  x3, 
fourth  term  of  sq.  rt.  =  [4  x7  -  2 (  -  5  x4)  (2  x3)]  -  2 (4  x5)  =  3  x2, 
last  term  of  sq.  rt.  =  V9  =  ±  3, 
next  to  last  term  of  sq.  rt.  =  —  36  x  -=-  2(  ±  3)  =  q=  6  x, 
third  from  last  term  of  sq.  rt.  =  [54  x2  -  (  T  6  x)2]  -*■  2(  ±  3  x)  =  ±  3 

This  term  (±  3x2)  is  numerically  the  same  as  the  fourth,  and  to  make  it 
the  same  in  sign  the  upper  signs  must  be  used.  Hence  the  square  root  is 
4  x5  -  5  x*  +  2  x3  +  3  x2  -  6  x  +  3. 

30.  49  a;10  4  14  a;9 -27  a;8 -60  a:7 

-  46  x«  4  80  x5  +  38  x4  +  4  a?  -  31  a?  -  30  a?  +  25. 

31.  16a10-16a9-52^  +  52a7 

4-  29  a,*6  - 102  a;5  +  55x4  + 106  a3  -  47  x2  +  16  a  +  64. 

32.  9  a:12 -30  a;11  4  25  a;10 -24  a?8  4  52  a;7 

-  50  xG  +  50  a,-5  +  16  x4  -  16  Xs  +  44  a* -20  x  +  25. 

33.  36  a;12  -  48  x11  -8x10  +  28  x9  +  56  x8  -  80  a;7 

-  79  a;6  +  78  ar>  4  47  x4  -  44  ar>  -  61  x2  +  42  a;  +  49. 

34.  49  x14  +  70  x13y  -  17  x12y2  -  44  ^y  4  111  xwy4  4  142  afy" 

-  35  a^/6  +  24  xhf  + 128  afy8  4  32  afy9  -  28  afy10 
+  80  x*yn  4  52  arfy12  -  24  a,y3  4  36  y14. 

Sug.  The  signs  of  the  last  four  terms  can  be  determined  by  noting  what 
sign  of  the  fifth  term  of  the  root  must  be  used  in  producing  the  fifth  term  of 
the  given  polynomial. 

CUBE    ROOT   OF   POLYNOMIALS 

Case  I 
171.    When  the  cube  is  a  quadrinomial. 

Rule.  Arrange  with  reference  to  one  letter  and  take  the  algebraic 
sum  of  the  cube  roots  of  the  extreme  terms. 

Dem.     By  the  binomial  formula   or  by  actual  multiplication 

we  have 

(a  ±  b)3  =  a3±3  a2b  4  3  ab2  ±  b3, 


CUBE  BOOT  OF  POLYNOMIALS  109 

in  which  it  is  seen  that  the  extreme  terms  of  the  quadrinomial 
are  the  cubes  of  the  respective  terms  of  the  cube  root. 

Test.  If  the  quadrinomial  is  a  perfect  cube,  its  second  term 
will  be  three  times  the  square  of  the  first  term  of  the  root  multi- 
plied by  the  second,  and  its  third  term  will  be  three  times  the 
first  term  of  the  root  multiplied  by  the  square  of  the  second. 

EXAMPLES   L 

Find  the  cube  root  of  each  of  the  following : 
1.   8aj»  -36 3*  +  5±x  -27.. 
Solution.     If  this  is  a  perfect  cube, 

first  term  of  cu.  rt.  =  V8x3  =  2  x, 
second  term  of  cu.  rt.  =  V  -  27  =  —  3. 
Hence  the  cube  root  is  2  x  —  3. 
Test.     We  observe  that 

3(2z)2(-3)  =  -36xV 
and  3(2x)(-3)2  =  54x. 

2.  64x6  +  96afy  +  48ay  +  82/3. 

3.  216  a* +  540  a?  +  450  x  +  125. 

4.  125^-75^y  +  15^-?/6. 

5.  64  x3  -  144  x*y2  +  108  xy4  -  21  y\ 

Case  II 

172.  When  the  cube  contains  only  three  powers  of  any  one  of  its 
letters. 

Rule.  Arrange  with  reference  to  any  one  of  the  letters  which  has 
only  three  powers,  regard  the  terms  not  containing  that  letter  as  the 
fourth  term  of  a  quadrinomial,  and  proceed  as  in  the  preceding 
case. 

EXAMPLES  LI 

Find  the  cube  root  of  each  of  the  following : 

l.   8  a6  -  36  a4b  +  54  a2b2  -27b3  +  48  aV  -  144  a2b<?  +  108  6V 
+  96  a2c4  -  144  be4  +  64  c6. 


110  HIGHER  ALGEBRA 

Solution.  Here  each  letter  has  but  three  powers.  Arranging  with 
reference  to  a,  and,  at  the  same  time,  arranging  with  reference  to  b  the 
terms  that  do  not  contain  a,  we  have 

8 a6  +  ( -  36  b  +  48  c2)  a4  +  (54  b2  -  144  be2  +  96  c4)  a2  +  (-  27  63 

+  108&2c2-144  6c1  +  64c6). 

Since  this  is  a  quadrinomial,  if  it  is  a  perfect  cube, 

first  term  of  cu.  rt.  =  V8  «6  =  2  a2, 


second  term  of  cu.  rt.  =  V  -  27  &3  +  108  62c2  -  144  6c*  +  64  c6 
=  _  3  b  +  4  c2. 

As  the  intermediate  terms  conform  to  the  test  named  in  Case  I,  the  given 
polynomial  is  a  perfect  cube,  and  the  cube  root  is  2  a2  —  3  b  +  4  c2. 

2.  a3  4  63+  c3+  3  a2b  +  3  a2c  4  3  ab2+  3  oc«+  6  abc  4  3  62c  +  3  6c2. 

3.  27  or3  4  54  afy  -  27  a2z2+  36  xy2-  36  a^-J-  9  xzA  +  8  y5  -  12  fz2 

4  6  2/z4  —  z6. 

4.  61  x6  -  96  x4f  +  144  x4z  4  48  arty6  -  144  tftfz  -f  108  arz2  -  8  */9 

+  362/6z-54^22  +  27z3. 

5.  125  w3  +  150  z<;2a;  -  225  wrfy  -  300  w2z  +  60  wx2  -  180  ivxy 

-  240  tans  +  135  wy2+  360 t<$2  4  240  wz2+  8  ar3-  36  a*y 

-  48  a&  4-  54«?/2  +  144  ^z  4  96  a*2  -27  f-  108  #2z 
-144^2-64z3. 

Case  III 
173.    Any  perfect  cube  not  included  in  Cases  land  II. 
Solution.     By  Art.  161  we  have,  after  arranging  in  descend- 
ing powers  of  x, 

(axn  4  bxn~l  4  cxn~2  4  dxn~3  4  •  •  -)3 
=  aV"4-3  a^-2+(3  a6243  a2c)  3*-24-(&8+6  a&c43a2cT)ar}w-34  •••• 

In  cubing  a  polynomial  of  the  nth.  degree,  a  term  in  ar3*  can 
result  only  from  the  cube  of  the  first  term  of  the  root ;  hence,  if 
we  take  the  cube  root  of  the  first  term  of  the  arranged  cube,  we  shall 
have  the  first  term  of  the  cube  root. 


CUBE   BOOT  OF  POLYNOMIALS  111 

A  term  in  a,*3"-1  can  result  only  from  three  times  the  square 
of  the  first  term  of  the  root  by  the  second ;  hence,  if  we  divide 
the  second  term  of  the  arranged  cube  by  three  times  the  square  of 
the  first  term  of  the  root,  ice  shall  obtain  the  second  term  of  the 
root. 

Terms  in  ic3"-2  can  result  only  from  three  times  the  first  term 
of  the  root  by  the  square  of  the  second  and  three  times  the 
square  of  the  first  by  the  third;  hence,  if  we  subtract  from  the 
third  term  of  the  arranged  cube  three  times  the  first  term  of  the  root 
by  the  square  of  the  second  and  divide  the  remainder  by  three  times 
the  square  of  the  first  term  of  the  root,  ice  shall  obtain  the  third  term 
of  the  root. 

Terms  in  a3"-3  can  result  only  from  the  cube  of  the  second 
term  of  the  root,  six  times  the  product  of  the  first  three  terms, 
and  three  times  the  square  of  the  first  term  by  the  fourth ;  hence, 
if  we  subtract  from  the  fourth  term  of  the  arranged  cube  the  cube  of 
the  second  term  of  the  root  and  six  times  the  product  of  the  first 
three  terms  of  the  root,  and  then  divide  the  remainder  by  three  times 
the  square  of  the  first  term  of  the  root,  we  shall  obtain  the  fourth 
term  of  the  root. 

In  the  same  way  four  terms  of  the  root  may  be  obtained  from  the 
other  end  of  the  cube,  i.e.  obtained  as  they  would  be  if  the  cube  were 
written  the  other  end  foremost. 

In  practice,  it  is  better  to  obtain  terms  alternately  from  the  two 
ends,  continuing  the  operation  only  until  the  terms  meet  in  the 
middle  of  the  root. 

174.  Sch.  1.  There  are  the  same  kinds  of  exceptions  here  as 
noted  in  Sch.  1  of  Case  III  for  square  root. 

175.  Sch.  2.  Test.  To  ascertain  whether  the  given  poly- 
nomial is  a  perfect  cube,  it  is  necessary  simply  to  note  how 
those  terms  not  used  in  the  operation  are  made  up  from  the  cube 
of  the  supposed  cube  root. 

176.  Note.  It  is  to  be  observed  that,  as  in  square  root  of 
polynomials,  the  cube  root  can  be  written  at  once,  without  writing 
any  intermediate  steps. 


112  HIGHER  ALGEBBA 

EXAMPLES  LH 
Find  the  cube  root  of  each  of  the  following : 

1.  64  x&  +  96  x5  -  96  x4  -  136  x?  +  72  x2  +  54  x  -  27. 
Solution.     If  this  is  a  perfect  cube, 

first  term  of  cu.  rt.  =  ^(34  x*>  =  4  x2, 
second  term  of  cu.  rt.  =  96  x5  h-  3(4  x2)2  =  2x, 
last  term  of  cu.  rt.  =  \/—  27  =  —  3. 

There  can  be  no  other  terms  in  the  cube  root,  as  there  can  be  no  terms 
between  2  x  and  —  3.  Moreover,  next  to  the  last  term  of  the  cube  root  is 
54  z  -f-  3(—  3)2  =  2x,  the  same  as  already  obtained  for  the  second  term. 
Hence  the  cube  root  is  4  x2  +  2  x  —  3. 

Test.  In  the  operation  all  the  terms  of  the  given  polynomial  have  been 
used  except  the  third,  the  fourth,  and  the  fifth.  By  reference  to  Art.  161  it 
will  be  seen  that  in  cubing  4  x2  +  2  x  —  3,  terms  in  x4  can  result  only  from 
three  times  the  square  of  the  first  into  the  third  and  three  times  the  square 
of  the  second  into  the  first,  giving 

3(4  x2)2(-  3)  +  3(2  x)2(4  x2)  =  -  96  x4. 

Terms  in  x3  can  result  only  from  the  cube  of  the  second  and  six  times  the 
product  of  the  three,  giving 

(2  x)3  +  6(4  x2)  (2  x)(-  3)  =  -  136  x3. 

Terms  in  x2  can  result  only  from  three  times  the  square  of  the  second  into 
the  third  and  three  times  the  square  of  the  third  into  the  first,  giving 

3(2x)2(-  8)  + 80-  3)2(4  x2)=  72  x2. 

Hence  the  polynomial  is  a  perfect  cube. 

2.  8  x6  -  36  x*y  +  114  a*tf  -  207  a?f  +  285  x2y4  -  225  xtf  + 125  f. 

3.  8  x12  -  36  xw  +  66  Xs  -  63  x«  +  33  xA  -  9  x2  +  1. 

4.  8  -  12a  +  42 x2 -  61  tf+  87 a** - 105^  +  87a)6-  66 x7  +  36^ 

Solution.    If  this  is  a  perfect  cube, 

first  term  of  cu.  rt.  =  \/8  ==  2, 
second  term  of  cu.  rt.  =  —  12  x  ■*■  3(2)2  =  —  x, 
last  term  of  cu.  rt.  =  \/—  8  cc9  =  —  2  xs, 
next  to  last  term  of  cu.  rt.  =  36  x8  -=-  3(-  2  x3)2  =  3  x2. 


CUBE  ROOT  OF  POLYNOMIALS  113 

As  there  can  be  no  other  powers  of  x  between  the  extreme  terms  2  and 

-  2  x3,  there  are  but  four  terms,  and  the  cube  root  is  2  —  x  +  3  x'2  —  2  x3. 

5.  8  x9  -  36  afy  +  66  afy2  -  87  ./'//'  +  105  x  y4  -  87  afy5  +  61  afy6  - 
42  afy7  -+- 12  a;?/8  —  8  ?/9. 

6.  64x9-96«8  +  192a;74-88x6-96ari  +  366^  +  147^-15ar,+ 
225  a; +  125. 

7.  x12  -  6x11  +  21  a10  -  47a9  +  81  a8  -  108  a7  +  126  zG  -  117  a5  + 
99  x4  -  61  ar>  +  42  x2  -  12  x  +  8. 

Solution.     If  this  is  a  perfect  cube, 

first  term  of  cu.  rt.  =  \/x^  =  x4, 

second  term  of  cu.  rt.  =  —  6  a;11  -h  3(x4)2  =  —  2  x3, 

third  term  of  cu.  rt.  =  [21  x10  -  3(x*)(-  2  x3)2]  +  3(x*)2  =  3 x2, 

last  term  of  cu.  rt.  =  V8  =  2, 

next  to  last  term  of  cu.  rt.  =  —  12  x  -^  3(2)2  =  —  x. 

There  can  be  no  other  powers  of  x  between  the  extreme  terms  x*  and  2. 
Moreover,  the  third  term  from  the  last  is  [42  x2  -  3(2)  (-  x)2]  +  3(2)2  =  3  x2, 
the  same  as  already  obtained  for  the  third  term  from  the  other  end.  Hence 
the  cube  root  is  x*  —  2  x3  +  3  x2  —  x  +  2. 

8.  64z12  +  144a;11  + 12 xw- 261  x?  +  66 Xs  +  387^  +  82^-  171a5 
+  666 a4  +  513a*5-  54.x2  -  324a;  +  216. 

9.  27 x15  -  54 x14  +  63 xl*  +  64 x'2  -  204 xn  +  87  xw  + 187  a9-414 Xs 

-  36  x7  +  433  x6  -  192  x5  -  192  x4  +  408  x?  +  165  x2  -  225  x  -  125. 

EXAMPLES  LIII 
Solve  the  following  by  the  principle  of  Art.  164  : 

1.  The  4th  root  of 

16  a4  -  96  a?x  +  216  tfx2  -  216  ax3  +  81  x\ 

2.  The  6th  root  of 

a6  _  6  a5b  +  15  a4b2  -  20  a3b3  +  15  a2b4  -6ab5  +  b6. 

3.  The  6th  root  of 

729  -  2916  x2  +  4860  x4  -  4320  x6  +  2160  a8  -  576  a10  +  64  x'2. 

4.  The  8th  root  of  x3  -  16  x7y  +  112  afy2  -  448  afy3  +  1120  afy4 

-  1792  afy5  +  1792  afy6  -  1024  a?y7  +  256  y8. 

downey's  alg.  —  8 


114  HIGHER  ALGEBRA 

177.  Prob.      To  extract  any  root  of  any  quantity. 

Solution.  By  the  Binomial  Theorem  or  by  actual  multiplica- 
tion we  have 

(a  +  &)2=a2  +  (2a  +  6)6, 

(a  +  bf  =  a3  +  (3  a2  +  3  ab  +  62)&, 

(a  +  6)5  =  a5  +  (5a4  +  10a86  +  10  a262  +  5  a&3  +  b4)  b, 

etc.,  etc.,  etc. 

In  reversing  the  process  it  is  seen  that  the  required  root  of  the 
first  term  of  the  given  polynomial  (or  of  the  first  period  of  the 
given  number)  will  be  the  first  term  (or  figure)  of  the  root. 

Could  we  divide  the  rest  of  the  polynomial  (or  number)  by  the 
part  in  the  parenthesis,  we  should  obtain  b,  the  next  term  (or 
figure)  of  the  root.  Only  the  first  term  (or  figure)  of  this  divisor 
is  known  until  b  is  found.  In  case  of  polynomials,  however,  only 
the  first  term  of  the  divisor  is  needed  (and  in  case  of  numbers 
this  first  figure  is  much  the  larger  part,  since  it  is  tens  with  ref- 
erence to  b  as  units).  After  the  second  term  (or  figure)  of  the 
root  is  found,  the  divisor  can  be  completed. 

If  the  root  contains  more  than  two  terms  (or  figures),  we  may 
form  a  new  trial  divisor  by  considering  the  whole  of  the  root  now 
found  as  the  first  term  (or  figure),  and  proceed  as  before. 

It  is  assumed  that  the  student  is  familiar  with  the  reason,  as 
given  in  Arithmetic,  for  pointing  off  the  number  into  periods  of 
as  many  figures  each  as  indicated  by  the  index  of  the  root,  begin- 
ning at  the  decimal  point. 

178.  Note.  As  already  shown,  the  square  root  and  cube  root 
of  polynomials  can  be  written  at  once  by  inspection.  In  the 
extraction  of  roots  of  numbers  the  work  is  greatly  facilitated  by 
the  use  of  logarithms,  as  shown  in  a  subsequent  part  of  the  work. 


CHAPTER    X 

SURDS  AND    IMAGINARIES 

SECTION  I  — SURDS 

179.  A  Radical  is  an  indicated  root  of  a  quantity. 

180.  A  Surd,  or  Irrational  Quantity,  is  a  radical  quantity  whose 
indicated  root  cannot  be  exactly  extracted. 

181.  A  surd  is  Quadratic,  or  of  the  Second  Degree,  Cubic,  or  of 
the  Tliird  Degree,  Biquadratic,  or  of  the  Fourth  Degree,  etc., 
according  as  its  index  is  2,  3,  4,  etc. 

182.  Similar  Surds  are  expressions  containing  the  same  surd 
factor. 

Thus,  2v/3a,  5 m V3 a,  (a2  —  m2)  VSa  are  similar  surds. 

183.  To  Rationalize  an  Expression  is  to  free  it  from  surds. 

184.  The  Rationalizing  Factor  is  the  factor  by  which  a  surd 
must  be  multiplied  to  rationalize  it. 

185.  To  Rationalize  the  Denominator  of  a  Fraction  is  to  free 
the  denominator  from  surds  without  changing  the  value  of  the 
fraction. 

186.  The  Simplest  Form  of  a  Surd  is  the  form  in  which  the 
smallest  possible  integral  number  is  left  under  the  radical  sign. 

187.  To  Simplify  a  Surd  is  to  reduce  it  to  its  simplest  form. 

REDUCTION  OF  SURDS 

188.  The  substance  of  all  demonstrations  in  Reductions  is  to 
show: 

1st.  That  the  operation  does  not  change  the  value  of  the  ex- 
pression. 

2d.   That  the  operation  produces  the  required  form. 

115 


116 


HIGHER   ALGEBRA 


189.    Prob.      To  simplify  a  surd  when  the  quantity  under  the  rad- 
ical sign  contains  a  factor  of  which  the  indicated  root  can  be  taken. 

Rule.     Write  the  required  root  of  this  factor  as  the  coefficient  of 
the  indicated  root  of  the  other  factor. 

This  is  because  the  root  of  the  product  equals  the  product  of 
the  roots  (Art.  150). 

EXAMPLES  LIV 
Simplify  the  following : 


1.   5aVl47ajy. 
Operation.     5  a  Vl47  x2y5  =  5  a  V49xV  x  '3y  =a  35  axy2  VS  y. 


2.    V32. 
5.  2V98. 


3.   V75. 
6.   3VI25. 


4.   V72. 
7.  4V180. 


a   V48a^3. 


9..  V63a5b\ 


10.   2cVl25aW. 


11.   4V108ay. 


14.    V56a:W. 


12.   3aV245  6V. 
15.   3VS!^y4. 


13.   6aV200 


m°x*. 


16.   5V135a&V. 


17.   aV432cW\      18.   5  6  V128  a4ra63"+6.     19.   7  V  -108  a667c9. 


20.   2V343ajy. 
23.   Va3  -f  a2b. 


21.   3 V720  a26-V.         22.   5V448tt36"6c7. 


25.    V20  a56'2  -  32  a6b3. 


27.   V3^--6^y  +  3^*. 


24.    Vl8a?-27afy. 

26.    V2x4  +  4^  +  2icV 


28.    V24a4-32a36. 


190.  Prob.  To  simplify  a  surd  when  the  quantity  under  the  rad- 
ical sign  is  a  perfect  power  of  the  degree  indicated  by  a  factor  of  the 
index. 

Rule.  Extract  the  root  indicated  by  this  factor  of  the  index  and 
write  the  result  as  a  surd  whose  index  is  the  other  factor. 

This  rule  is  simply  an  application  of  Art.  150. 


SEDUCTION  OF  SURDS  117 

EXAMPLES  LV 

Simplify  the  following : 

1.  4343. 

Operation.  -^343  =  V\^H  =  V7. 

2.  </49.  3.   </169.  4.   4225  a4&2. 
5.    4'49.                          6.    ^216.  7.    4l000  a*b6. 
8.    481.                          9.    a/625  asb4.                10.    464. 

11.   4-  729  aW..       12.    41728  a3wi+6.  13.   2^/32  a5b~w. 

14.   4l96a664.  15.   SK/Sl^b8.  16.   4729  a6&«. 

17.    4l6a^  +  64»2/  +  64  2/2. 


18.   V64  x3  -  576  x2y  +  1728  as^2  —  1728/. 

191.  Prob.     To  reduce  a  rational  quantity  to  the  radical  form. 

Rule.  Raise  the  quantity  to  the  degree  of  the  required  radical 
and  place  the  result  under  the  corresponding  radical  sign. 

As  the  operation  performed  is  the  inverse  of  the  operation 
indicated,  the  value  of  the  quantity  is  not  changed. 

192.  Prob.  To  introduce  under  the  radical  sign  the  coefficient  of 
a  surd. 

Rule.  Raise  it  to  the  power  of  the  same  degree  as  the  surd  and 
multiply  the  result  by  the  quantity  under  the  radical  sign. 

Dem.  The  coefficient  being  reduced  by  the  last  rule  to  the 
same  form  as  the  surd,  we  have  the  product  of  two  like  roots, 
which  is  equal  to  the  same  root  of  the  product  (Art.  150). 

EXAMPLES  LVI 

1.  Reduce  to  radicals  of  the  2d  degree, 

3  ab2,  7  xhf,  11  xy\  and  2  x  -  3  f. 

2.  Reduce  to  radicals  of  the  3d  degree, 

4  aPy,  5  a¥,  —  3  a~2bc2,  and  6  xy2z~3. 


118  HIGHER  ALGEBRA 

In  the  following  introduce  the  coefficients  under  the  radical 
sign  or  within  the  parenthesis : 

3.   3V2.  4.   4«V#a.  5.   2VJ. 

6.   \  VS.  7.   3^/3.  a   ^25. 


f  \        a?J  "   m  — »  >m  +  a 


15.    '*{*-?)*.  16-    ^(l-J)-  17-    (a-1)V5+1- 


193.  Prob.  To  reduce  surds  of  different  degrees  to  equivalent 
ones  of  the  same  degree. 

First  Rule.  Write  for  the  common  index  the  1.  c.  m.  of  all  the 
indices,  and  raise  each  quantity  under  the  radical  sign  to  a  power 
ivhose  degree  is  the  factor  by  which  its  root  index  must  be  multiplied 
to  produce  this  1.  c.  m. 

Second  Rule.  Write  the  surds  with  fractional  exponents,  reduce 
these  exponents  to  a  common  denominator,  and  express  the  results  in 
the  radical  form. 

Dem.  By  either  rule  the  value  of  each  surd  remains  the  same, 
since  we  extract  a  certain  root  and  then  raise  the  surd  to  the 
corresponding  power. 

194.  Sch.  Surds  are  readily  compared  in  magnitude  by  first 
reducing  them  to  equivalent  surds  of  the  same  degree. 

EXAMPLES   LVII 

Reduce  the  following  to  equivalent  surds  of  the  same  degree : 
1.    V3#,  -y/2x2,  and  Vox:{. 

OPERATION 

1/bx*  =  ^(SajS)*  =  v"126jR 


REDUCTION   OF  SURDS  119 

Or  we  may  proceed  thus  : 

y/2x*  =  2^  x$  =  2&  x&  =  y/Wx\ 
■  W&  =  &x*  =  5T2  x&  =  Wl6?. 
2.   V2  and  -^3.  3.    1/2  and  ^/2. 

4.    V2  and  V^.  5.    y/t<?  and  \/3i, 

6.    V3  and  W.  7.    #5*  and  ^&» 

8.    Va,  "v^6,  and  Vc.  9.    V3a,  V2  6,  and  V6c. 

10.   3,  VS,  and  </Il.  11.   a3,  W,  and  #?. 

12.   Va  +  6  and  V«  —  b.  13.   V2  a,  V2  a,  and  V2  a. 

14.  avfaaF*  2^^,  and  4-y/12¥3. 

15.  5\/l0¥3,  7^4^,  and  6^7a-y. 
Determine  which  is  greater, 

16.    V7    or  Vl8.  17.    ^2    or  </3. 

18.    V3    or  </l6.  19.    ^5    or  </l4. 

20.    V&3  or  V#4.  21.    -y/x*  or  V^5- 

195.  While  the  value  of  a  numerical  surd  cannot  be  determined 
exactly,  it  may  be  determined  to  any  required  degree  of  accuracy 
by  the  use  of  decimal  places  in  the  root.  To  find  the  approxi- 
mate numerical  value  of  V7  -*-  V5  we  may  proceed  thus : 

^I^2-645751 -  =  1.183215-. 

V5     2.236068  ••• 

This,  however,  involves  three  tedious  operations,  viz.,  extracting 
the  square  root  of  7,  extracting  the  square  root  of  5,  and  divid- 
ing the  first  result  by  the  second.  By  multiplying  numerator 
and  denominator  by  V5,  the  second  of  these  operations  may  be 
avoided,  and  the  third  replaced  by  a  much  shorter  one.     Thus, 

V7  =  V7  x  V5=  V35  =  5.916079  -  _11832i5  .... 
V5  '  V5      V5        5  5 


120  HIGHER   ALGEBRA 

This  suggests  as  suitable  form  for  all  surd  fractions  equivalent 
fractions  with  rational  denominators  (Arts.  178  and  180). 

196.  Conjugate  Surds  are  quadratic  surds  which  differ  only  in 
the  sign  of  one  of  their  terms. 

Thus,  Va  +  V&  and  Va  —  Vb,  a  +  V&  and  a  -  Vb,  Va  +  b  and  Va  —  b, 
Va  +  Vb  +  Vc  and  Va  +  Vb  —  Vc,  etc.,  are  general  forms  of  conjugate 
surds. 

197.  Prob.     To  rationalize  the  surd  denominator  of  a  fraction. 
Rule.     1st.    WJien  it  contains  a  monomial  quadratic  surd,  multi- 
ply both  terms  of  the  fraction  by  this  surd. 

2d.  When  it  contains  a  monomial  surd  of  any  other  degree,  as 
-\/xm,  in  which  m  <n,  multiply  both  terms  of  the  fraction  by  -\/xn~m. 

3d.  When  it  contains  a  binomial  quadratic  surd,  multiply  both 
terms  of  the  fraction  by  the  conjugate  siird. 

4th.  When  it  contains  a  trinomial  quadratic  surd,  multiply  both 
terms  of  the  fraction  by  one  of  the  surds  conjugate  to  this,  and  then 
by  the  surd  conjugate  to  the  reduced  product  of  these  two  surds. 

Dem.     1st.    Va  X  Va  =  a,  a  rational  quantity. 

2d.    -\/xm  x  -Vxn~m  =  -\/xn  =  x,  a  rational  quantity. 

3d.    ( Va  ±  V&)  ( Va  =F  Vb)  =  a  —  b,  a  rational  quantity. 
Also     (a  ±  Vo)  (a  T  V&)  =  a2  —b,  a  rational  quantity. 

4th.  ( Va  ±  V6  ±  Vc)  ( Va  ±  V&  T  Vc)  =  ( Va  ±  Vb)2  -  c 
=  a  +  b  —  c  ±  2  Va6 ;    and  (a  +  b  —  c  ±  2  Va6)  (a  +  b  —  cT  2VaFj 
=  (a  +  6  —  c)2  —  4  ab,  a  rational  quantity. 

EXAMPLES   LVIH 

Reduce  the  following  to  equivalent  forms  having  rational 
denominators : 

1.     3Vg 


4V7a 

±VTx     *VYx"  VTx        28x    '   "  28  x 


n                         3V5         3V5    vV7x     3\/35x           3       ,^— 
Operation.      = x = ,  or  V35  x. 


REDUCTION  OF  SURDS 


121 


2. 


6  x 


5</3a? 


Operation. 
4 


<; 


6*     x^  =  6^  =  2^  ^ 


3. 


V3 

3V2 
2V3* 


5v/3x2     5S!/3x2      v/32x 
1 

V2* 
6V6ft 


15  x 


2Vft2  +  ft,y  +  y2 
3  Vft  —  # 

12.  uVf 

15.  8VJ. 


18. 


21. 


3  V3  x 

2l/2x*' 

Sab 


v  27  a4 


22.    18  \7- 


9a 
16/ 


25.     *r* 


28. 


V(2-ft)2 
5 


V5-V2 
Operation.  — 


4. 


7x 


2V5ft 
8 


10. 


7vTz 

aVx2  —  xy  +  y'2 


b~vx-t-y 

13.   20aJp-- 
*5ft 


16. 


8  « V4~ft 


7  6^/2^ 
19.   7</f. 

Sug.     Write  in  the  form 


Vft^2 
11.   3V|. 

</3 
17.    ^| 

20.    12\/|. 
3  aft 


V3%* 


23. 


16  ft 


^64^ 

26.    ^+y° 


V(ft  +  2/)3 


24. 


27. 


1-ft2 
VWft2" 
ft4  —  ?/4 


29. 


31. 


33 


3-V2 

5+V3 
5-V3* 

Vs2  +  1  -  g; 
Vft2  + 1  +  ft 


5__= 5 xv^±^=51^5_±V?l=5CV5+V2) 

V5     V2     V5-V2     V5+V2  6-3 

30.    ^£±^S 

Vft— Vy 


32. 


34. 


VI  +  ft2  —  ft 
Vft  +  l+Vg  —  1 
Vft  + 1  —  Vft  —  1 


122  HIGHER  ALGEBRA 


__    Vx2  +  x  +  Var*  —  x  „    cc2  H-  a  4-  xVa?  +  a 

35.    — ^^^2 »  3d.    ^r:^^^ • 

V&2  4-  5  —  V&2  —  a  as  +  Va,*2  -f  a 

37  .T  +  l+V^TT  ^  a  +  ^-vV  +  frV 

x  +  l-Va^-l  a-ta  +  Vc^  +  fc2^ 

39.  — ■* 

Operation.      S x  V5  + V5  +  Vg  =  8(V6  + V5  + V§) 

V5  +  V3  -  V2  V5  +  V3  +  V2              3  +  >/15 
and 

6(V5+V3+V2)  x  3-V15  =  6(-2V3  +  3V2-v/30)  =  ^30  +  2  V5-3  V2. 

3+VT5              3-VI5  9~15 

8  1 

40. 41. 


V3+V2  +  1  2V2+V3+V5 

198.  Prob.  To  find  the  factor  which  will  rationalize  any  given 
binomial  surd. 

Solution.  The  general  form  of  such  surd  being  y/cf  ±  V&*, 
we  have, 

r  «  nr  m»  1  1 

Vo"r  ±  Vb1  =  a*  ±  6"  =  a™  ±  b™  =  (a™)™1  ±  bms)™. 

Now,  with  the  lower  sign  in  all  cases  and  with  the  upper  sign 
when  mn  is  even,  anr  —  bms,  which  is  rational,  is  divisible   by 

(anr)mn  ±  (bms)mn  (Art.  99).     When  mn  is  odd,  anr  +  b™,  which  is 

rational,  is  divisible  by  (anr)mn  +  (bms)mn  (Art.  99).  In  either  case 
the  quotient,  which  can  be  written  by  forms  (1)  and  (2),  Art.  99, 
will  be  the  rationalizing  factor. 

EXAMPLES  LIX 

Find  the  rationalizing  factor  of  each  of  the  following : 

1.    V3+^5. 

Operation.     V3  +  y/E  =  3*  +  5*  =  3^  +  5^  =  (33)  *  +  (52)  K 

Now  33  —  52  is  the  difference  of  the  same  even  powers  (the  6th  powers)  of 

the  terms  of  (8*)^ +<#)*,  or  3^  +  5*  and  hence  is  divisible  by  8*4-6* 
(Art.  99).     The  quotient,  which  may  be  written  by  form  (2),  Art.  99,  is  the 


ADDITION  AND   SUBTRACTION  OF  SURDS  123 

factor  by  which  32  -f  5*,  or  >/3  +  \/b,  must  be  multiplied  to  produce  33  —  52, 
or  2,  a  rational  quantity. 

(33  _  52)  -  (3*  +  5*) 

=  (3*)*  -  (3*)* (5^)  +  (3*)3 (5*)2 _ (3I)2 (51)8  +  (3i) (5i)4  _ (51)6 

=  VJP  -  9^5  +  V3«  #P  -  15  +  V3  v/5*  -  #P. 
2.    ^3-^2.  3.   2+^4. 

4.    V2-^5.  5.    </?-$?. 

ADDITION  AND   SUBTRACTION  OF   SURDS 

199.    Prob.      To  add  or  subtract  surds. 

Rule.  Reduce  each  surd  to  its  simplest  form;  then  to  the  alge- 
braic sum  of  the  coefficients  of  the  similar  surds  annex  the  common 
surd  part,  and  indicate  the  addition  or  subtraction  of  the  dissimilar 
surds. 

This  is  but  an  application  of  the  principles  of  Arts.  29  and  40. 

EXAMPLES  LX 
Perform  the  following  indicated  combinations : 

1.  V27+3VI08  +  2V75. 

Operation.  V27  =    V9  x  3    =    3V3 

3VT08  =  3V36  x  3  =  18 y/S 

2V75    =2V25o^  =  10V3 

Ans.,  31v/3 

2.  3V50  +  4V72.  3.  2V75  +  3V245. 
4.   5  V147  +  4  Vi08.  5.   G  V175 -f  7  V252. 
6.  4V150-3V96.  7.   5V396-V176. 

8.    VT08  +  3  V27—  2  V48.  9.   |Vl80  +  ^V80- |Vi25. 

10.   V54  +  3v/l28.  11.   5^108-3^32. 

12.   3  6^/128  a4b2  -  a^/Wab1.       13.   3  xJ/M2lc  +  5</l62^. 


124  HIGHER  ALGEBRA 

14.  bVWrf  +  xVMlrf -4^162^. 

15.  7aV2SS~as-7a2-\/512a3. 

16.  3^/486^  +  2-^/64^. 

17.  13^324¥4-2V338aT4. 

18.  2V|  +  3V|+V^. 
Operation.  2Vf  =  4\/|  =  4Vf  =  $v/6 

V^  =  i  Vf  =  f  Vf  =|V6 
19.   6V|  +  7V24.  20.   2V|  +  |V240. 

-  ^+#-#     -  ^i-i2vi. 

23.  Vf  +  Vi  +  i-v/9.  24.  Vfl  +  Vj  +  fv/sI. 

25.    Vf+VS-Vf.  26.    Vf+V9-Vf|. 

27.  3Vl47-jVi-2V^.  28.   18^  +  -v/243-6^/I 


29.   7  6  V18  a4  -  27  a262  -  2  a  V8  a262  -  12  b\ 


30.    6V8a36+l6a4  +  aV27fc4  +  54a&3. 


31.   « =  + 


a  -f-  Va2^^2     a  —  Va2  —  a2 


32    V  a  +  #  +  Va  —  &      Va  +  a;-Va-a; 
Va  +  #  —  Va-«      Va  +  a?  +  s/a  —  x 

33.   //a/3+V2V|  /V3-V2V  2 

W3-V2/      W3+V2/      5-2V6* 


MULTIPLICATION  OF  SURDS  125 

MULTIPLICATION  OF   SURDS 

200.  Prob.      To  multiply  surds. 

Rule.  If  the  surds  are  of  different  degree,  reduce  them  to  equiva- 
lent surds  of  the  same  degree;  then  to  the  product  of  the  coefficients 
annex  the  common  root  of  the  product  of  the  quantities  under  the 
radical  sign,  and  reduce  the  result  to  its  simplest  form. 

This  is  but  an  application  of  the  principles  of  Arts.  193  and  148. 

201.  Sch.  1.  If  the  quantities  under  the  radical  sign  are 
integral,  the  surds  should  first  be  simplified  if  not  already  in  the 
simplest  form,  as  the  factors  that  may  be  removed  are  much  more 
readily  detected  before  the  multiplication  than  after.  Thus,  we 
may  write, 

V28  x  V72  =  V2016  =  Vl44  x  14  =  12  VH ; 

but  after  multiplying,  considerable  inspection  is  required  to  find 
the  largest  square  factor,  and  the  difficulty  is  greater  with  larger 
numbers.     We  should  proceed  thus  : 

V28  x  V72  =  2  V7  x  6  V2  =  12  Vl4. 

202.  Sch.  2.  When  the  quantities  under  the  radical  sign  are 
fractions,  it  is  usually  best  not  to  simplify,  but  to  employ  cancel- 
lation as  far  as  possible.     Thus, 

Vf  x  V|  x  V^  =  V|  x  |  x  V-  =  V3. 

203.  Sch.  3.  Much  work  is  often  saved  by  resolving  the 
quantities  under  the  radical  sign  into  such  factors  as  will  show 
those  that  are  common,  even  when  the  surds  are  in  the  simplest 
form,  as  this  reveals  any  factors  that  may  be  removed  after  the 
multiplication.     Thus,  we  may  write, 

V42  xV77=V3234=V49  x66  =  7V66; 
but  much  inspection  is  saved  by  writing 

V42  xV77=Vinr6  xV7  x  11  =  V72  x  66  =  7  V66. 
Again, 


T05xV70=V7x5x3xV7x5x2=V72x52x6  =  35V6. 


126  HIGHER  ALGEBRA 

204.  Sch.  4.  If  multiplicand  or  multiplier  or  each  of  them 
is  a  polynomial,  the  sum  of  the  partial  products  must  be  taken, 
as  in  the  multiplication  of  rational  polynomials. 


EXAMPLES  LXI 
Perform  the  following  multiplications : 

1.    V3  by  Vl2.  2.  V2l  by  VT. 

3.    V20  by  VS.  4.  V9  by  y/S. 

5.    -V36x>  by  J/6x.  6.  3Vl5  by  V50. 

7.   4V6  by  3V12.  8.  Jjxy-  by  tyify. 

9.   4^/36  by   V48.  10.  2VH  by  V2l. 

11.    V60  by  V30.  12.  V30  by  V35. 

13.   7V35  by  V65.  14.  V39  by  V91. 

15.    V231  by  VIM.  16.  V9l  by  V182. 

17.   S/|  by  Vf.  18.  Vf  by  Vf 


19.   5V24a26  by  7&V32«. 

Solution.  As  the  product  of  the  same  roots  equals  the  root  of  the  prod- 
uct (Art.  148),  if  these  surds  are  first  reduced  to  the  same  degree,  the  multi- 
plication can  be  performed. 

Now,  5^24  a*b  =  5^8  x  3a*b  -  10y/S¥b  =  lOVWaW 

1bVWa  =  lbVlQx2a  =  28bV2a  =  2$by/8<? 


Product,  280  ab\Zl2ab* 

20.  4V2  by  5V3.  21.  2V24  by  3^3. 

22.  </2tfb  by  Vl6^6.  23.  Vl2  by  y/3. 

24.  V3^  by  V27^.  25.  V7  x  V2  x  </E. 

26.  VjxVfxVV-- 

r,  3/T  /2         6/27         6/T        6/23         6/33       . 

Operation.     V2X\5XA/Y  =  ^X%xV|=1- 


MULTIPLICATION   OF  SURDS  127 

27.    Vf  by  3\/|.  28.   6V^  by   \/2. 

29.    72x^X^X^1.  30.    Vf  X  a/I  X  </f . 

31.    VT+x  by  -v/l-ha?.  32.   Va— 6 x  Va+fc X  Va*11^. 

33.   3V6-4V3  +  3  by  2V6  +  5V3. 

Operation.  3\/6  -  4>/3  +  3 

2V6  +  5V3 


36-24V2+6V6 
-60  +  45V2  +  15v/3 

24+21\/2  +  6V6+  15V3 

34.  5-2V3  by  4  +  3V3. 

35.  2V^  +  3V2  by  6V^-V2. 

36.  V5+V3-V2  by  V5.+  V2. 

37.  5  +  3V2  by  5-3V2. 

Sue     In  multiplying  the  sum  of  two  quantities  by  the  difference,  as  in 
the  37th,  always  apply  the  principle  of  Art.  63. 

3a  3V5  +  2V3  by  3V5-2V3. 

39.  2VII-5  by  2V14  +  5. 

40.  7a^-4aV3a  by  7a*  +  4aV3a. 

41.  V5-V3-V2  by  V5  +  V3-V2. 

42.  V6  +  V3+V5  by  V5  +  V3-V6. 

43.  5V8  +  6VI2-2V20  by  7V2-3V3 +  4V5. 

44.  3a-3Vab  +  2b  by  3a  +  3Vab  +  2  6. 

Sec     In  solving  examples  like  the  last  four,  the  terms  should  be  so 
grouped  as  to  give  the  product  of  the  sum  and  difference  of  two  quantities. 


45.  3Va  +  Va-9a  by  3Va-V*-9a. 

46.  V&--Vxy  +  -y/tfby  Vx  +  Vy. 

47.  ^5-2^6  by  3^4-^36. 


128  HIGHER  ALGEBRA 

DIVISION  OF  SURDS 

205.  Prob.     To  divide  surds. 

Rule.  If  the  surds  are  of  different  degree,  reduce  them  to  equiva- 
lent  surds  of  the  same  degree;  then  to  the  quotient  of  the  coefficients 
annex  the  common  root  of  the  quotient  of  the  quantities  under  the 
radical  sign,  arid  reduce  the  result  to  its  simplest  form. 

This  is  but  an  application  of  the  principles  of  Arts.  193  and  150. 

206.  Sch.  When  the  division  gives  rise  to  a  fraction  with 
a  surd  in  the  denominator,  this  denominator  should  be  rational- 
ized. 

EXAMPLES   LXII 

Perform  the  following  divisions : 

1.  18V6  by  3V10. 

Operation.  ^VS  =  gjjf  =  Q    13  =  6  ^ 

3V10         >10         >5      5 

2.  8  a  V6o"  by  ±J/Ta~2. 


Operation.     ^^  =  2«:^  =  2a^  =  2^B4lF. 

3.   6V12  by  3V3.  4.  10V15  by  2.VK. 

5.   V32  by  V8.  6.  V42  by  V84. 

7.   ^80^2  by  ^5^.  8.  a2</l92  by  aVs. 

9.    V^T?  by  V^+^.  10.  a/(^-3?/)3  by  V(^-32/)3. 

11.   V30  by  V42.  12.  Vl5  by  S^15. 

13.   6  by  ^6.  14.  -^81  a2  by  3  a. 

15.   VlO  by  v®.  16.  V2  by   ^8. 

17.   -\/^  by  a^.  18.  aWl2^6  by  ab2J/2lab. 

19.    V|  by  VJf.  20.  Vfi  by  Vff. 

21.    ^  ty  ^t-  22  5-VI35  by  V5. 


INVOLUTION   OF  8UBDS  123 

23.  GV32-12V24  +  3V5  by  3V2. 

24.  Vl4  -  V35  by  V2  -  V5. 

25.  3-^3+^3  by  VS. 

26.  2  V3  -f  7  V2  by  5  V3  -  4  V2. 

Bug.     Write  as  a  fraction  and  rationalize  the  denominator. 

27.  5V5  +  3V2  by  V5  +  V5. 

28.  a  +  b  —  c  +  2  Va£  by  Va  +  y/h  —  Vc. 

30.    V5-J-2V6-V5-2V6  by  V5 +  2V6 +V5 -2V6. 

INVOLUTION   OF   SURDS 

207.  Prob.     To  raise  a  monomial  surd  to  any  power. 

Rule.  Either  divide  the  index  of  the  root  by  the  exponent  of  the 
power,  or  raise  the  quantity  under  the  radical  sign  to  the  required 
power;  then  annex  the  result  to  the  required  power  of  the  coefficient, 
and  reduce  to  the  simplest  form. 

This  is  but  an  application  of  the  principles  of  Arts.  148 
and  151. 

208.  Cor.  A  surd  is  raised  to  a  power  whose  exponent  is  the 
index  of  the  root  by  simply  removing  the  radical  sign. 

EXAMPLES   LXIH 

Perform  the  following  indicated  operations : 

1.   (6V3*)*.  2.   (-6Vp)2.  3.  (±y~5xhjY. 

4.   (SV-2ab2)5.  5.   (-iVEaf.  6.  (Sy/Uay. 

7.    (-Z/SSaWf.  8.    (2^8^)4.  9.  (</2aWy. 

10.    (-s/laWf.  11.    (^50^/)4.  12.  {^V-21diW)\ 

13.   (^64  afy2)3.        14.    (V7-V5)2.         15.  (3+V6)2. 

16.    (V3-V6)2.  17.    (^2--\/3)3.  18.    (^3+s!/2)3. 

downey's  alg.  —  9 


130  HIGHER  ALGEBRA 

EVOLUTION   OF    SURDS 
209.    Prob.      To  extract  any  root  of  a  monomial  surd. 

.  Rule.  Either  extract  the  required  root  of  the  quantity  tinder  the 
radical  sign,  or  multiply  the  index  of  the  given  root  by  the  index  of 
the  required  root,  then  annex  the  result  to  the  required  root  of  the 
coefficient,  and  reduce  to  the  simplest  form. 

This   is    but   an   application   of   the   principles   of   Arts.    148 
and  151. 

EXAMPLES   LXIV 
Perform  the  following  indicated  operations : 
1.    -yJ{VaVf).  2.    V(36V257?^). 

3.    */(-  27^04oW).  4.    -*/(awh5J/3X a'0M). 


5.    -ty(8V7ax).  6.    V(^1024*¥> 

7.    ^a~Va.  8.    vWsi/2. 

9.    .£/(43  xy*V43xf).  10.    Vx\x-y)5n. 


11.    V(v49-70^  +  25x4).  12.    ^/(64«%']V64a:i63); 

210.  Theorem.  The  square  root  of  a  binomial,  one  of  ivhose 
terms  is  rational  and  the  other  a  quadratic  surd,  can  be  found  when- 
ever the  rational  term  is  separable  into  two  parts  the  product  of  whose 
square  roots  is  half  of  the  surd  term. 

Dem.  Let  a±b-\/c  be  such  that  a=m-\-n,  and  ±-J-&Vc=Vmn. 
Then  a  ±  bVc  =  m  +  n  ±  2^/mn  =  (Vm  ±  Vw)2.  Therefore  the 
square  root  is  Vm  ±  y/n. 

EXAMPLES  LXV 
Extract  the  square  root  of  each  of  the  following : 

1.   27-10V2. 

Solution.  27  -  10 V2  =  25  -  10 v2  +  2  =  (5  - \/2)2.  Hence  the  square 
root  is  5  —  V2.  Of  course,  the  negative  of  this  rQot,  viz.,  V2  —  5,  is  equally 
admissible. 


IMAGI NAMES  131 


2.   12-2V35. 


Solution.  12  -  2  V35  =  7-2  V35  +  5  =  (  VI  -  V5)2.  Hence  the  square 
root  is  V7  —  V5. 

3.    4  +  2V3.  4.    3  +  2V2.  5.    7-2V10. 

6.   8-4V3.  7.   30  +  10V5.  8.   8  -  2V15. 

9.   70 -30  VS. 

Solution.  70  -  30>/5  =  25  -  30  V5  +  9  x  5  =  (5  -  3\/5)2.  Hence  the 
square  root  is  5  —  3  V5. 

10.  18  +  8V5.  11.   11-4V6.  12.  49  +  12V5. 

211.  Note.  The  method  of  inspection  here  given  is  of  limited 
application.  A  general  method,  applicable  to  all  perfect  squares 
of  the  kind  mentioned  in  the  theorem,  requiring,  however,  the 
solution  of  two  equations,  one  of  the  first  and  the  other  of  the 
second  degree,  will  be  given  as  an  application  of  the  process  of 
solving  such  equations. 

SECTION  II  — IMAGINARY   QUANTITIES 

212.  An  Imaginary  Quantity  is  an  expression  which  contains  an 
indicated  even  root  of  a  negative  quantity. 

Thus,  V- 1,  V- x2,  \/— 12,  3±2v/^~4,  a  ±  bV—  1  are  imaginary 
quantities. 

213.  All  quantities  not  imaginary  are  called  Real  Quantities. 

214.  Expressions  which  contain  both  real  and  imaginary  terms 
are  often  called  Complex  Quantities. 

215.  When  the  radical  sign  affects  a  negative  quantity,  we  can- 
not regard  it  as  indicating  a  possible  arithmetical  operation.  For 
example,  V—  9  is  neither  +  3  nor  —  3,  for  neither  multiplied  by 
itself,  i.e.  squared,  will  produce  —  9,  but  +  9  instead.  Since  im- 
aginary quantities  occur  frequently  in  mathematical  investigations 
and  lead  to  important  results,  we  need  to  consider  what  meaning 


132  HIGHER   ALGEBRA 

should  be  attached  to  them  in  order  that  they  may  obey  the 
ordinary  laws  of  algebra. 

The  expression  Va  is  understood  to  be  such  that  Vax  Va=a. 
Now  if  we  agree  that  V—  «  is  such  that,  in  the  same  way, 
V—  axV-  a  =  —  a,  we  shall  find  that  imaginary  quantities 
obey  the  algebraic  laws  already  established.  This  agreement 
would  forbid  our  first  multiplying  together  the  quantities  under 
the  radical  sign,  observing  the  law  of  signs,  and  then  extracting 
the  square  root ;  for  in  that  ease  we  would  have  V—  a  X  V—  a 
=  Va2  =  ±  a.  The  agreement  limits  us  to  the  minus  sign,  as  it 
should,  since  the  square  root  of  a  quantity  multiplied  by  itself 
should  produce  the  original  quantity. 

216.  Theorem.  Every  monomial  imaginary  can  be  reduced  to  the 
form  b^s/—  1,  in  which  b  may  be  either  rational  or  surd. 

Dem.  The  general  form  of  a  monomial  imaginary  is  mV-  k, 
in  which  n  is  even.  Now,  since  the  root  of  the  product  equals  the 
product  of  the  roots,  mV- k=m-\/k(— l)  =  m~\/k-\/— l  =  b-\/ —  1, 
where  b  =  m-\/Tc. 

217.  Theorem.  Every  'polynomial  containing  some  real  terms  and 
some  imaginary  terms  of  the  same  degree  can  be  reduced  to  the  form 
a  ±  b^J  —  1,  in  which  a  and  b  may  be  either  rational  or  surd. 

Dem.  This  is  evident  from  the  fact  that  all  the  real  terms  can 
be  combined  into  one  (it  may  be  a  polynomial)  and  represented  by 
a,  and  the  imaginary  terms  into  another  represented  by  ±  b~V—  1. 

218.  Cor.  The  general  form  of  an  imaginary  of  the  second  degree 
is  a  ±  b  V—  1,  in  which  b  is  rational  or  surd,  and  a  is  rational, 
surd,  or  0. 

219.  The  Imaginary  Element  is  some  even  root  of  —  1  which 
renders  an  expression  imaginary. 

Thus,  in  V^x  =  VxV^l,  bV^x2  =  5 x\/^T, 

2  +  V^4  =  2+  2V"^T,    V^S2  =  2v^2  =  2\/2v/^T, 
the  imaginary  elements  are  \/—  1,  V—  1,  y/—  1. 

V—  1,  sometimes  called  the  Imaginary  Unit,  is  often  represented  by  i. 


IMAGINARIES  133 

220.  Theorem.  When  the  index  of  the  imaginary  element  is  a 
composite  number  and  one  of  tfie  factors  odd,  the  value  is  not  changed 
by  rejecting  the  odd  factor. 

Dem.  In  the  form  bV  —  1  let  the  factors  of  n  be  p  and  q, 
q  being  odd.  Then  b V^T  =  b *V-1  =  b^V^l  =  bV^l,  since 
any  odd  root  of  —  1  is  —  1 . 


Thus,  3v^nr  =  3Vv/^l  =  3v^T, 


221.  Conjugate  Imaginaries  are  imaginaries  of  the  second  degree 
which  differ  only  in  the  sign  of  the  imaginary  part. 

Thus,  a  +  frV^HE  and  a  -  6V— 1,  5  +  3v^l  and  5  -  SV-T,  mV^l 

and  —  mV—  1,  4V—  1  and  —  4V—  1,  etc.,  are  conjugate  imaginaries. 

222.  Theorem.  Both  the  sum  and  the  product  of  two  conjugate 
imaginaries  are  real. 

Dem.  b  V^l  +  (-  6V^T)  =  0, 

and  a+bV^T  +  a  —  bV^l  =  2a. 

Also  bV^l  x  (-  &V^T)  =  -  b\-  1)=  62, 

and  (a  +  6 V^l)  X  (a  -  &V-T)  =  a2  +  b2, 

the   last  being  the  product  of  the  sum  and  difference   of  two 
quantities,  which  is  the  difference  of  their  squares. 

223.  The  Modulus  of  a  quadratic  imaginary  of  the  form  a±&V— 1 
is  the  positive  square  root  of  a2  +  b2. 

Each  of  the  imaginaries  that  constitute  a  conjugate  pair  has 
the  same  modulus,  and  this  common  modulus  is  seen  to  be  the 
positive  square  root  of  the  product  of  the  two. 

224.  Prob.     To  add  or  subtract  imaginaries. 

Rule.  Reduce  each  to  the  form  in  which  the  imaginary  element 
is  expressed  by  the  smallest,  possible  index  (Art.  220).     If  the  imagi- 


134  HIGHER  ALGEBRA 

nary  element  is  then  the  same  in  each,  combine  its  coefficients  and 
write  the  result  as  the  coefficient  of  the  common  imaginary  element; 
if  the  imaginary  element  is  not  the  same  in  each,  indicate  the  addi- 
tion or  subtraction. 

•  If  the  imaginary  element  is  the  same  in  each,  the  problem  is 
simply  that  of  uniting  quantities  of  the  same  kind. 

EXAMPLES  LXVI 

Combine  the  following : 

1.  6+V:r4  and  S+V^. 
Operation.  6  +  v^Ti  =  6  +  2\/^T 

Sum,  14  +  5V^T 

2.  2-^8  and  5a/^2. 


Operation.     2v^8  =  2  Vv^8  =  2V-  2  =  2v/2v/~^T 


Sum,  7V2V-1 


3.  4V-r27  and  6V-32. 

Operation.     4 V- 27  =  4 V9  x  3  x(-  1)  =  12 Vs V^T, 
6V-32  =  6Vl6  x  2  x(-  l)=24V2v/^T. 

Here,  although  the  imaginary  element  is  the  same  in  both  quantities, 
they  contain  dissimilar  surds.  Hence  we  cannot  unite  into  one  term,  but 
have 

12V3V"^I  +  24\/2V^Tr=  12(V3  +  2V2)\/^T. 

4.  5  a  +  3  6  V11^  and  3  a  +  2  6  V11!. 


5.   15V- 16  and  GV"^.         6.   V- 225  and  V-169. 


7.   y-9a2  and  V-4ar>.         a  3 V- 9  and  -V-49. 
9.   SV"17!^  and  2V^48.       10.   V-289  and  -V-U. 


11.   aV-8  and  V-18a2.       12.   7+-4V-27  and  3-2V-12. 
13.    \^Tl6  and   -J/~^l.         14.   4  +  2^/^  and  2  -  </^a. 

15.  ty^32  and  V^8. 

16.  V^il-v^T ^-V^12l  +  ^-243. 


IM AGIN  ARIES 


135 


225.    Prob.      To   determine   the   different  powers   of  V—  1  and 

</- 1.  . 

Solution.  (V—  l)l=  V—  1, 

(V-l)2=-1,  by  Art.  215, 

( V=T)3=  ( V^T)2  V^T  =  -  V^T, 
(V=3)*=[(V^T)T     =1, 

(V:=:T)5=(V-i)V:^T  =  V- 1, 

(V^T)fi=  (V^T)V^T  =  - 1, 
( V^T)7=( V=T)6V==T  =  -  V-  i , 

(V=1)8=[(V=I)<]*       =h 
etc. 

It  is  thus  seen  that  we  have  a  succession  of  the  results,  V—  1, 
_  i?  —  V—  1,  and  1,  or  that  the  only  forms  assumed  by  the 
different  powers  of  V— 1  are  ±V— 1  and   T  1. 

(^/^T)3= (</3i)*,J/:ri  =  v=T</- 1, 

(^)W1, 

(</^)"=(</^T)^^l  =  -</- 1, 
(<^T)«=  (</=Iy(l/=lf=  -  V^T, 
(</- 1)'= (^/- l)  6^=~1=  -  V=1</^T, 

(^-l)"=[«/^l)4J=l- 
etc. 

It  is  thus  seen  that  Ihe  only  forms  assumed  by  the  different 
powers  of  \^^1  are  ±  \'r— T,  ±  V— T,  ±  V^T^^T,  and  T  1. 


226.    Prob.      To  multiply  or  divide  imaginaries. 

Rule.  Reduce  to  the  form  in  which  the  root  index  of  the  imagi- 
nary element  is  the  same  in  each  imaginary  term',  and  multiply  or 
divide  as  in  case  of  other  quantities,  observing  (he  principles  of 
Art.  225  for  the  product  or  quotient  of  the  imaginary  elements. 


136 


HIGHER  ALGEBRA 


EC  AMPLE  3   LXVII 
Multiply  the  following : 

1.   4V^  by  2V^. 
Operation*.  4  V^3  =  4  V3  V^T 

2V^2  =  2V2V^1 


2.   7\/-32  by  5</~- 


Operation. 


Product,     8  V6  (  -  1 )  =  -  8  V6. 

'^2. 

"32  =  7  </l6\/^2  --=  14  ^2  v>^l 

5V^2=    5^'2'v/^l 


Product,     1QV2(V-  l)2, 
or    70V2V^T 


3.  V—  a,*2  by  V— ?/2 


5.  _2V-25  by  3V-36. 


4.  5  V— 5  by  4V  — 3. 

6.  V^2  by  V^8. 
8.  3^/^Yl  by  4vcr3. 


7.  V-196  by  V-27. 

9.  UV^  +  V^bj^V^l.  10.  4-j-2V^4  by  5  -  3 V^I. 

11.  1  +  V^l  by  1-V-X  12.  12+3  V^  by  12-3 V^f. 
13.  3-J/^9  by  4^~8. 


Operation. 


3#-  9  =  3J/qV^1  =  3Vi 


5  4/ 


4^/_  8  =  4\/8\/-  1  =  4V2 V-  1  =  4 v/2(\/-~l)* 


Product,     12V0(v-  l)3=12v/6V-l  V-l. 
14.  3a/^2  by  5V^.  15.  3a/^2  by  5^3. 

Find  the  value  of  each  of  the  following : 
16.  (5V^2)2.  17.  (2V^3)3.  18.  (4^2)4, 

19.  (3V^2)5.  20.  (2^2)6.  21.  (V^7)7. 

Find  the  modulus  of  each  of  the  following : 
22.  V2+V^2.  23.  V2-V^2.  24.  3  +  2\/^3. 

25.  3_2V^3.  26.  4-3V:=l.  27.  4  +  3V:=:l. 


MAG  I  NAMES 


137 


Kationalize  the  denominators  of  the  following : 


28. 


30 


3-V=2 
2 


V-9 

Divide  the  following: 
32.   6V-16  by  2Vz:4. 

GV^nTi      24 


29. 


31. 


Operation. 


4V-  1 


2V-4 

The  imaginary  elements  cancel. 

33.   2V:rT  by    v^2. 

noc.»  ™™       2\/^T     2(\/^T)^     2</^T 
Operation.      =s— ? —  = 


V32+2V-3 

V"-^2-2V^3* 

3V^2  +  2V^5 

3V:r2-2V^5 


=  6. 


=  ^8v^l=^T8. 


y/~2        ^2C/-~i  ^2 

34.   16V^"l)  by  2 v7^4.  35.  tV^i  by  2V3& 

36.    1  by   V—I.  37.   36  by  12V:=rI. 


38.   63V- 16  by  V-81. 

40.   4+V^2  by  2-V^2. 


39.   V-16  by  V-4. 


Operation.  When  there  is  more  than  one  term  In  the  divisor,  it  is  usually 
best  to  write  the  quotient  as  a  fraction  and  then  rationalize  the  denominator. 
Thus, 

4+>/z~2=z4  +V2>/3T_(4+V2Vin)(2  +  V2V^-"1) 
2  _  V^2      2  -  V2  V^T  4  +  2 


6 


41.   1  +  V^l  by  1  -V—  1. 

43.  a+V-«  by  a  —  V—  #• 

44.  s-i+^Zlby   - 


42.   1  by  3— 2V:=~3. 


45.    Simplify  a+V-ft^a 


2z-l+V-3 
^6 


a  —  V—b      a  +  V— 6 


138  HIGHER   ALGEBRA 

Extract  the  square  root  of  each  of  the  following : 

46.  5  +  12V^T. 

Operation.     5  +  12Vj-l  =  9  +  12v^l  -  4  =  (3  +  2>/^~l)2.     Hence  the 
square  root  is  8  -f  2V—  1. 

47.  -5  +  12V-1. 

Sua.     Write  in  the  form  4  +  1SV^-T  -  9. 

48.  _7_24V^T. 

Sug.     Write  in  the  form  9  -  24  V^l  -  16. 

49.  7  +  12V^4.  50.  21-20V:::T 
51.  -9  +  20V:=:l.  52.  35-12V::=T. 
53.  40  +  42V:=rI.  54.  33-56V^l. 
55.   27  +  36V:=:T.  56.   24-70\/^l. 

57.    Simplify  —-£ —  4- 


58.    Simplify 


2-5V-1     2+5V-1 

V3  —  a;  +  V^2      V^^x  —  V^5 


227.  In  Algebra  imaginary  quantities  have  their  greatest  prac- 
tical use  in  showing  that,  arithmetically,  the  conditions  of  prob- 
lems in  the  solution  of  which  they  occur  cannot  be  fulfilled.  For 
example,  let  it  be  required  to  divide  10  into  two  parts  whose  prod- 
uct shall  be  40.  Proceeding  in  the  usual  way  we  find  the  two 
parts  to  be  5  -f- V—  15  and  5  —V—  15.  These  imaginary  results 
show  that  the  requirements  of  the  problem  cannot,  in  the  arith- 
metical sense,  be  fulfilled.  It  is  easy  to  show  that  the  largest 
product  of  any  two  parts  of  10  is  25.  Nevertheless,  these  imagi- 
nary values  do  satisfy  the  algebraic  requirements  of  the  problem, 
the  sum  of  5  +  V—  15  and  5  —V—  15  being  10  and  the  product 
40.  Imaginary  quantities  occur  as  well,  and  always  in  conjugate 
pairs,  as  roots  of  an  equation  which  also  has  one  or  more  real 
roots. 


IMAGTNARIES  139 

Imaginaries  also  have  a  practical  use  in  Algebra  in  showing  the 
limits  of  a  variable  quantity ;  for  example,  in  finding,  by  algebraic 
methods,  the  maximum  or  minimum  value  of  a  function. 

In  Analytical  Geoinetry  imaginaries  have  a  frequent  and  im- 
portant use  in  showing  the  limits  of  loci  represented  by  certain 
forms  of  equations. 

In  a  branch  of  Mathematics  known  as  Quaternions  imaginary 
quantities  have  a  graphical  and  definite  signification.  If  any 
magnitude  be  represented  by  a,  then  —  a  represents  an  equal 
magnitude  in  the  opposite  direction.  Now  ax(— 1)  =  — a; 
hence  —  1,  regarded  as  an  operator,  causes  a  reversal.  We  have 
seen  that  «xV-l  X  V—  1  =  —  a;  hence  V—  1,  regarded  as  an 
operator,  causes  a  reversal  when  repeated. 

Now  any  magnitude  may  be  represented  graphically  by  a  length 
laid  off  on  a  straight  line.  Suppose  this  length  to  be  a.  Then 
—  a  would  be  the  same  length  in  the  opposite  direction.  Now  we 
may,  if  we  choose,  regard  V—  1  as  being  an  operator  which  turns 
a  through  a  right  angle,  because,  when  repeated  in  the  same  direc- 
tion, it  causes  a  reversal.  Thus,  just  as  a  x  V— 1  x  V—  1  =  —  a, 
a  turned  successively  through  two  right  angles  gives  —  a ;  and 
V—  1,  represented  by  i  in  Quaternions,  is  the  operator  which, 
when  repeated,  causes  this  reversal. 


CHAPTER  XI 
SIMPLE   EQUATIONS 

SECTION  I  — SIMPLE   EQUATIONS   WITH   ONE   UNKNOWN 

QUANTITY 

228.  An  Equation  is  an  assertion  by  means  of  a  mathematical 
symbol  that  two  expressions  have  the  same  value.  This  symbol 
is  the  Sign  of  Equality  (Art.  10). 

229.  The  Members  of  an  Equation  are  the  expressions  connected 
by  the  sign  of  equality,  the  one  on  the  left  being  called  the  First 
Member  and  the  one  on  the  right  the  Second  Member. 

230.  An  Identical  Equation,  called  also  an  Identity,  is  an  equa- 
tion that  is  true  for  all  values  of  the  letter  or  letters  involved.* 

Thus,  (x  +  a)  (x  —  a)  =  x2  —  a2  is  an  identity. 

231.  An  Equation  of  Condition,  usually  called  simply  an  Equa- 
tion, is  an  equation  that  is  true  only  for  a  limited  number  of 
values  of  the  letter  or  letters  involved. 

Thus,  x  +  5  =  9  is  true  only  f or  *  =  4  ;  and  x2  —  6  x  =  7  is  true  only  for 
x  =  7  and  x  —  —  1. 

232.  The  Unknown  Quantity  is  the  letter  whose  value  is  re- 
quired. 

233.  An  Absolute  Term  of  an  equation  is  a  term  which  does  not 
contain  an  unknown  quantity. 

234.  A  Numerical  Equation  is  one  in  which  the  known  quantities 
are  represented  by  numbers. 

*  Some  writers  use  the  sign  =  to  indicate  that  two  expressions  are  identi- 
cally equal.  The  sign  is  not  used  in  this  work,  as  the  distinction  is  not 
thought  to  be  of  sufficient  importance. 

140 


SIMPLE  EQUATIONS  141 

235.  A  Literal  Equation  is  one  in  which  some  or  all  of  the  known 
quantities  are  represented  by  letters. 

236.  The  Degree  of  an  Equation  is  the  same  as  that  of  its  term 
of  highest  degree  (Art.  50),  the  unknown  quantity  or  quantities 
first  being  freed  from  fractional  and  negative  exponents. 

Thus,  ox2  +  bx  =  c  is  of  the  second  degree,  3 x2y  —  bxy +  2x  +  6y2=  12  is 

of  the  third  degree,  xy  =  m  is  of  the  second  degree,  -  +  -  =  5  (which,  by  mul- 

x     y 

tiplying  both  members  by  xy,  becomes  y  +  x  —  5  xy)  is  of  the  second  degree, 

x2  +  3  x  —  5  x-2  —  3  =  0  (which,  by  multiplying  both  members  by  x'2,  becomes 

x*  4-  3  x8  —  5  —3  x2  =  0)  is  of  the  fourth  degree. 

237.  A  Simple  or  Linear  Equation  is  one  of  the  first  degree. 

238.  Equations  of  one  unknown  quantity  are  called  Quadratic, 
Cubic,  Biquadratic,  or  Quintic,  according  as  they  are  of  the  second, 
third,  fourth,  or  fifth  degree. 

239.  Any  equation  above  the  second  degree  is  called  a  Higher 
Equation. 

240.  A  Root  of  an  Equation  is  a  value  of  the  unknown  quantity 
which  renders  the  equation  true. 

241.  To  solve  an  Equation  is  to  find  its  root  or  roots. 

242.  An  equation  is  said  to  be  Satisfied,  and  a  supposed  root 
Verified,  when,  on  substituting  for  the  unknown  quantity  the  sup- 
posed root,  the  equation  becomes  an  identity. 

243.  In  solving  equations  the  following  axioms  are  employed : 

1.  If  the  same  quantity  or  equal  quantities  be  added  to  or  sub- 
tracted from  both  members  of  an  equation,  the  equality  of  the  mem- 
bers will  not  be  destroyed. 

2.  If  both  members  of  an  equation  be  multiplied  or  divided  by 
any  quantity  that  is  not  equal  to  0,  the  equality  of  the  members  will 
not  be  destroyed* 

*  For  the  introduction  and  loss  of  roots  by  multiplying  or  dividing  by 
a  factor  containing  the  unknown  quantity,  see  Art.  351. 


142  HIGHER   ALGEBRA 

To  show  that  the  equality  of  the  members  may  be  destroyed 

by  dividing  both  members  by  a  quantity  that  is  equal  to  0,  let  us 

take  the  equation 

5x-15  =  2x-6,  (1) 

which  is  seen  to  be  satisfied  for  x  =  3. 
By  factoring  we  have 

5(x-3)  =  2(x-3).  (2) 

Now  if  we  should  divide  both  members  of  (2)  by  x  —  3,  we 
should  have  the  absurd  result  5  =  2.  The  two  members  of  (2) 
are  equal,  not  because  of  the  relation  of  5  to  2,  but  because 
x  —  3  =  0,  and  5  times  0  is  the  same  as  2  times  0.  In  dividing 
by  x—  3  we  remove  the  element  that  makes  the  two  members 
equal. 

244.  Prob.  To  solve  a  simple  equation  ivith  one  unknown  quan- 
tity. 

Rule.  1.  Clear  the  equation  of  fractions,  if  it  have  any,  by 
multiplying  each  term  by  the  1.  c.  m.  of  all  the  denominators. 

2.  Transpose  the  unknown  terms  to  the  first  member  and  the 
known  terms  to  the  second  member,  by  changing  the  sign  of  each 
term  transposed. 

3.  Unite  similar  terms,  and  divide  both  members  by  the  coefficient 
of  the  unknown  quantity. 

Dem.  1.  Multiplying  by  the  l.c.m.  of  all  the  denominators 
does  not  destroy  the  equality  of  the  members  (Art.  243,  2),  and 
it  clears  of  fractions  because  in  multiplying  any  one  of  the  frac- 
tions by  this  1.  c.  m.  the  denominator  of  that  term  is  canceled  by 
one  factor  of  the  multiplier. 

2.  When  a  term  is  dropped  from  one  member,  it  is  subtracted 
from  that  member.  Now  if  it  is  written  with  its  sign  changed  in 
the  other  member,  it  is  subtracted  from  that  member  also,  and 
the  equality  of  the  members  is  not  destroyed  (Art.  243,  1). 

3.  Dividing  by  the  coefficient  of  the  unknown  quantity,  after 
uniting  terms,  does  not  destroy  the  equality  of  the  members 
(Art.  243,  2),  and  it  leaves  in  the  first  member  simply  the  un- 
known quantity;  hence  the  second  member  is  its  value,  or  the 
root  of  the  equation. 


SIMPLE  EQUATIONS  143 

245.  Cor.  The  signs  of  all  the  terms  of  an  equation  may  be 
changed  without  destroying  the  equality  of  the  members. 

For  this  is  the  same  as  multiplying  or  dividing  by  —  1. 

246.  Sen.  Equations  having  terms  of  higher  degree  than  the 
first  often  reduce  to  simple  equations  by  the  disappearance  of 
these  terms  in  collecting. 

SOME   PRACTICAL   SUGGESTIONS 

247.  1.  When  there  are  several  integral  terms  and  but  few 
fractional  terms,  time  is  saved  by  collecting  terms  before  clear- 
ing of  fractions. 

2.  In  clearing  of  fractions  the  student  must  be  careful  to 
change  the  signs  of  the  terms  in  the  numerator  of  a  fraction  that 
is  preceded  by  the  minus  sign  (Art.  11). 

3.  In  simple  cases  the  terms  may  be^transposed  and  united  at 
the  same  time. 

4.  Factors  which  appear  in  both  members  of  the  equation 
should  be  canceled,  and  equal  terms  with  opposite  signs  in  the 
same  member  of  the  equation,  or  the  same  sign  in  opposite  mem- 
bers of  the  equation,  should  be  stricken  out. 

5.  When  a  fraction  has  a  polynomial  numerator  and  monomial 
denominator,  it  is  often  better  to  separate  the  fraction  into  parts 
by  dividing  each  term  of  the  numerator  separately  by  the 
denominator, 

6.  It  is  often  expedient  to  clear  of  the  smaller  or  simpler 
denominators  first,  and  after  each  step  to  see  that  by  transposi- 
tion, uniting  terms,  etc.,  the  equation  is  kept  in  as  simple  a  form 
as  possible. 

EXAMPLES  LXVIU 
Solve  the  following : 

.,     o      ,  2#-10     Sx     5a;-14 

1.  o.  x  H = 

3  2  4 

Solution.  Clearing  of  fractions  by  multiplying  every  term  by  12,  the 
1.  c.  m.  of  the  denominators,  we  have 

36  x  +  8  x  -  40  =  18  x  -  15  x  +  42. 


144  HIGHER   ALGEBRA 

Observe  that  the  minus  sign  before  the  last  term  of  the  given  equation 
denotes  that  the  whole  term  is  to  be  subtracted.  Hence  when  the  vinculum 
is  removed  in  clearing  of  fractions,  we  must  either  still  indicate  this  subtrac- 
tion by  writing  —  (15  x  —  42),  or  perforin  it  by  changing  the  signs. 

Transposing,        36  x  4-  8  x  —  18  x  -f  15  x  —  42  +  40. 

Collecting  terms,  41  x  —  82. 

Dividing  by  the  coefficient  of  x,  x  =  2. 

x  —  a_  (x  —  b)2 
2     ~  2x-a 

Solution.     Performing  the  indicated  involution  and  clearing  of  fractions, 

2  x°-  -  2  ax  -  ax  4  a2  =  2  x*  -  4  bx  +  2  62. 

dropping  2  x2  from  both  members,  transposing  and  uniting, 

4  bx  -  3  ax  =  2  62  -  a2, 

or  (4  6  -  3  a)x  =  2  62  -  a2. 

.       _  2  62  -  a2 
45  -3a' 

The  answer  would  be  equally  correct  if  written 

s  =  *-2*.     Why? 
3a-46 

3.   8a;-5(4ar+3)  =  25-8aj.  4.   5(aj-  2)  -  6  (a +  4)  =  21. 

5.    ^^  +  ^  =  20-^i5.  6.    2x-3x  +  7  =  -  +  l. 

11  ^ 

6  x  —  4      9  _  18  —  4  # 
3  3 

x  4- 1      X  —  1  4 


2      ' 

fr   ~ 

2 

3 

X  +  4: 

19x-3     4 

-7 

a? 

2 

8 

12 

2 

5            2 

SB 

-2 

SB  +  2      x2  - 

-4 

2 

a +  3 

_4x  +  5 

3 

x  —  4 

6a>-l 

a 

g 

-1     x-2_x-5 

ar- 

6 

10. 


35  —  1       £  -f-  1       X2  —  1 


I    6x+7     7a?-13=2o;+4 
9         6a?+3         3 

3x  2x       2ar}-5 


13.      v    XJiz=ac4-^.  14. 


2a;+3     2x-3     4a^-9 

.       2(a?-7)        a?-2^a;+< 
X-2     x-3     jc-6     x-7  *   x2+3x-28     a,--4     x+7 


16       2(»-7)    +x=2==x±3m 


17. 


SIMPLE  EQUATIONS  145 

2x  + 1       2x-l  _      9#  +  17 


2^-16     2x  +  12     a? -2  a;- 48 


„  „    a  —  b,b  —  c      a  —  c 

18. 1 = 

x — c     x—a        x 

19.  (a2  +  s)  (ft2  +  a?)  =  (aft  +  z)2. 

20.  (a'4-a)4-(x-a)4-8a.^  +  8a4  =  0. 

21.  (a?-l)3  +  .^  +  (z  +  l)3  =  3a;(a;2-l). 

22.  (x-l)(x  +  2)(x-3)  =  x2(x-2)  +  2(x  +  4). 

248.  Theorem.  7/*  a;  +  V#  =  a  +  Vft,  tot  itf/uc/i  a;  awrf  a  are 
rational  and  V#  and  Vft  «wrd,  ^e  rational  and  surd  terms  are 
separately  equal. 

Dem.     Transposing,  we  have 

x  —  a  =  Vft  —  Vy . 

Since  a  rational  quantity  cannot  equal  a  surd,  this  equation 
can  be  true  only  when 

x  —  a  =  0  and  Vft  —  V#  =  0, 
whence  #  =  a  and  Vft  =  V#. 

249.  Theorem.  7/"  x  +  V  —  #  =  a  -f  V—  ft,  wi  w/wc/i  a;  and  a  are 
rad  and  V— y  and  V—  ft  imaginary,  the  real  and  imaginary  terms 
are  separately  equal. 

Dem.     Transposing,  we  have 

#  —  a  =  V— ft  —  V  —  y. 

Since  a  real  quantity  cannot  equal  an  imaginary  one,  this  equa- 
tion can  be  true  only  when 

#  —  a  =  0  and  V  —  ft  —  V—  .y  =  0, 
whence  #  =  a  and  V—  6  =  V— 2/. 

250.  Many  equations  containing  surds  become  simple  equa- 
tions after  being  freed  from  surds  and  reduced.  No  general  rule 
can  be  given  for  solving  such  equations,  as  different  cases  must 

downey's  alg.  — 10 


146  HIGHER  ALGEBRA 

have  different  treatment.  Much  depends  on  the  student's  insight. 
When  inspection  does  not  suggest  what  steps  will  free  the  equa- 
tion from,  surds,  the  student  must  resort  to  trial,  being  careful  to 
preserve  the  equality  of  the  members.  We  give  here  a  few  sug- 
gestions. Some  of  the  operations,  as  will  be  shown  in  Art.  351, 
cause  roots  to  be  lost  or  extraneous  roots  to  be  introduced. 

1.  By  making  a  surd  term  constitute  one  member  of  an  equa- 
tion, its  radical  sign  will  disappear  when  both  members  are  raised 
to  a  power  of  the  same  degree  as  the  surd.  A  repetition  of  the 
process  (having  treated  the  most  complicated  surd  first  and 
reduced  as  much  as  possible)  will  cause  a  second  radical  sign  to 
disappear,  and  so  on. 

2.  When  a  surd  denominator  is  similar  to  the  surd  numerator 
of  another  fraction  or  to  another  term,  it  is  usually  best  to  multi- 
ply both  members  by  this  denominator. 

3.  It  is  sometimes  best  to  rationalize  a  surd  denominator,  espe- 
cially when  it  differs  only  by  a  sign  from  the  numerator  of  the 
same  fraction. 

4.  Sometimes  a  surd  factor,  or  a  factor  containing  a  surd  term, 
can  be  removed  either  from  both  terms  of  a  fraction  or  from  both 
members  of  the  equation. 

EXAMPLES  LXIX* 

Solve  the  following : 

1.    Va-32  =  16-Va.  2.   - =  c. 

3.    VS  +  V^7  21  6*-9 


-Vx  —  7  V  5  x  +  3 

5.    Vfl  -|- s/x  -f-  va  —  V#  =  Vx. 


6.  V(l  +  a)2  +  (1  -  a)x  +  V(l  -  a)2  +  (1  +  a)x  =  2a. 

7.  V113+V[7+V(3+V^)]S  =  4. 


I  Vl+v/3+V6«  =  2. 


*  Most  of  the  examples  of  this  set  are  from  Olney's  University  Algebra. 


SIMPLE  EQUATIONS  147 

9    V6x-2  =  ^6x-9 
VWx  +  2     ±V6x  +  6 

16  a;  — 3 


11    a  ~  ^a2  ~  ®*  —h  12      ax~  1    _  a  _j_  Voal 


a  -f-  Va2  —  x2  VaaJ-f-1  2 

13    3  Va;  -  4  =  15  +  3  Va;  Vox  -{-  V&  _  Va  +  V& 

2+Vx        40+Va;  Va«-V6  V& 

15    Va- vg-Vft2-aa; _  ,  Vm  +  Vra  —  y  _  t_# 

Va+^a--y/a2-ax  Vm-Vm-y     m 

17    a  +  ?  +^2  aa;  +  a;2  _  &2       18    Va;  +  1  —  Va;  —  1  _  1t 
a  +  ^-V^aaJ  +  aj"  '    VaT+T  +  -y/x^l     3 

19.    Va  +  x  +  Va  —  a;  =  &. 

2a    1  +  a? +  V2a? +  a?'=aV2  + a; +Vjp. 
1  +  x  —  V2aT+a?        V2  —  x  —  Va; 

^    V3a;  +  l4-V3a=1 
V3a;+1-V3a; 

22.    _I =  +  _=! -2& 

Va  —  a;  +  Va      Va~^le  —  Va        x 

23  a  +  2a;+Va2-4x2:=g^. 
tt  +  2a;-Va2-4arJ       a 

24  18(7a;-3)^250V2a;  +  l 

2x  +  l        '    3V7a;-3 


PROBLEMS   LEADING   TO   SIMPLE   EQUATIONS   WITH  ONE 
UNKNOWN  QUANTITY 

251.  The  solution  of  a  problem  by  algebraic  methods  consists 
of  two  distinct  parts:  1st.  Forming  the  equation  or  equations, 
which  consists  in  expressing  the  conditions  of  the  problem  in 
algebraic  language.     2d.  Solving  the  equation  or  equations. 


148  HIGHER   ALGEBRA 

252.  In  solving  an  algebraic  problem  involving  but  one  un- 
known quantity,  we  usually  proceed  as  follows : 

1.  We  represent  the  unknown  quantity  by  one  of  the  final 
letters  of  the  alphabet. 

2.  We  form  an  equation  by  indicating  the  operations  that 
would  be  necessary  to  verify  the  result  if  it  were  known. 

3.  We  solve  this  equation. 

253.  While  we  usually  represent  the  number  sought  by  x,  it  is 
sometimes  advantageous  to  represent  it  by  some  multiple  of  x. 
For  example,  if  three  numbers  whose  sum  is  36  are  in  the  ratio  of 
2,  3,  and  4,  we  may  avoid  fractions  by  letting  2x,  3x,  and  4  x  be 
the  numbers,  giving  the  equation  2  x  +  3  x  -f-  4  x  =  36.  In  other 
cases  it  may  be  advantageous  to  represent  by  x  some  number 
(time,  for  example)  from  which  the  number  sought  (distance,  for 
example)  is  readily  found. 

EXAMPLES  LXX 

1.  A  bicyclist  made  a  trip  at  the  rate  of  13  miles  an  hour.  His 
return  by  a  road  10  miles  longer  at  the  rate  of  16  miles  an  hour 
required  15  minutes  more  time.     What  was  the  first  distance  ? 

Solution.     Let  x  =  the  first  distance  ; 

then  x  +  10  =  the  second  distance, 

—  =  the  first  time, 
13 

and  £-E —  =  the  second  time. 

16 

Now,  since  this  second  time  is  longer  than  the  first  by  \  of  an  hour, 

x_  _  x  +  10  _  I 
13         16         4* 

Clearing  of  fractions,     16  x  =  13  x  +  130  -  52. 

Transposing  and  collecting,        3  x  =  78. 

Therefore  x  =  26. 

2.  A  can  do  in  20  days  a  piece  of  work  which  B  can  do  in 
12  days.  A  began  the  work,  but  after  a  time  B  took  his  place, 
and  the  whole  work  was  finished  in  14  days  from  the  beginning. 
How  long  did  each  work  ? 


SIMPLE  EQUATIONS  149 

Solution.    Let  x  —■  the  no.  of  days  A  worked  ; 
then  14  —  x  =  the  no.  of  days  B  worked  ; 

^V  =  the  part  of  the  work  A  did  in  1  day, 
^  =  the  part  of  the  work  B  did  in  1  day, 

—  =  the  part  of  the  work  A  did  in  x  days, 

14  —  x 

■ —  =  the  part  of  the  work  B  did  in  14  —  x  days. 

Now  the  sum  of  what  A  and  B  did  in  their  respective  times  was  the  whole 
work,  or  1.     Hence 

x_     14  —  x  _  , 
20         12  ' 

Clearing  of  fractions,        3  x  +  70  —  5  x  =  60. 

Transposing  and  collecting,  —  2  x  =  —  10. 

Therefore  x  =  5, 

and  14  -  x  =  9. 

3.  A  post  is  I  in  the  earth,  ^  in  the  water,  and  13  feet  in  the 
air.     Find  the  length  of  the  post. 

fit  D 

4.  A  post  is  —  in  the  earth,  —  in  the  water,  and  a  feet  in  the 

r  n  '  q 

air.     Find  the  length  of  the  post. 

5.  If  an  outward  trip  is  made  at  the  rate  of  30  miles  an 
hour,  the  return  trip  at  18  miles  an  hour,  and  the  whole  time  is 
10  hours,  what  is  the  distance  ? 

6.  When  a  is  taken  from  the  numerator  of  a  fraction  whose 
numerator  is  b  less  than  its  denominator,  the  value  of  the  fraction 

becomes  — .     What  is  the  original  fraction  ? 
n 

7.  When  5  is  taken  from  the  numerator  of  a  fraction  whose 
numerator  is  3  les§  than  its  denominator,  the  value  of  the  fraction 
becomes  f .     What  is  the  original  fraction  ? 

Sue  Let  a  —  5,  b  —  8,  m  =  3,  and  n  =  7,  and  substitute  in  the  formula 
obtained  from  the  solution  of  the  preceding  example.  We  thus  obtain  at 
once  from  the  general  case  the  result  for  the  special  case. 


150  BIG  HER   ALGEBRA 

8.  A  man  having  completed  f  of  his  journey,  finds  that  after 
traveling  30  miles  farther  only  f  of  the  journey  remains.  Find 
the  length  of  the  journey. 

9.  A  man  rows  down  a  river  for  2  hours  at  a  rate  that  would 
take  him  4  miles  an  hour  in  still  water ;  then,  resting,  he  floats 
with  the  current  for  half  an  hour.  He  then  rows  back  to  the 
starting  place  in  3  hours.     Find  the  rate  of  the  current. 

10.  A  grocer  drew  14  gallons  of  syrup  from  a  cask  which  had 
lost  J  part  by  leakage,  and  found  that  f  remained.  Find  the 
capacity  of  the  cask. 

11.  A  and  B  can  together  do  in  12  days  as  much  work  as  A  can 
do  alone  in  20  days.  In  how  many  days  could  B  alone  do  the 
same  amount  of  work  ? 

12.  A  cistern  can  be  filled  by  the  first  of  three  pipes  in  1J  hours, 
by  the  second  in  2\  hours,  and  by  the  third  in  5  hours.  In  what 
time  can  they  together  fill  the  cistern  ? 

13.  A  can  do  in  15  hours  a  piece  of  work  which  B  can  do  in 
25  hours.  After  A  has  worked  for  a  certain  time,  B  completes 
the  job,  working  9  hours  longer  than  A.  How  many  hours  did 
A  work  ? 

14.  A  cistern  which  contains  2400  gallons  can  be  filled  in 
15  minutes  by  three  pipes,  the  first  of  which  lets  in  10  gallons 
per  minute,  and  the  second  4  gallons  less  than  the  third.  How 
many  gallons  does  the  third  let  in  per  minute  ? 

15.  A  tank  can  be  filled  by  one  of  two  pipes  in  24  minutes, 
and  by  the  other  in  30  minutes,  and  emptied  by  a  third  in  20  min- 
utes.    In  what  time  will  the  tank  be  filled  if  all  are  left  open  ? 

16.  A  can  row  4  and  B  3  miles  an  hour  in  still  water.  A  is 
14  miles  farther  upstream  than  B,  and  they  row  toward  each 
other  till  they  meet,  4  miles  above  B's  starting  place.  Find  the 
rate  of  the  current.  • 

17.  On  a  sum  of  money  borrowed,  annual  interest  is  paid  at 
5%.  After  a  time  $200  is  paid  on  the  principal,  and  the  inter- 
est on  the  remainder  is  reduced  to  4%.  By  these  changes  the 
annual  interest  is  lessened  one  third.    What  is  the  sum  borrowed  ? 


SIMPLE  EQUATIONS  151 

18.  A  grocer  has  two  kinds  of  coffee,  one  at  a  cents,  and  the 
other  at  b  cents  a  pound.  How  much  of  each  must  he  take  to 
make  a  mixture  of  n  pounds  at  c  cents  a  pound  ? 

19.  A  grocer  has  two  kinds  of  tea,  one  at  GO  cents  and  the 
other  at  90  cents  a  pound.  How  much  of  each  must  he  take  to 
make  a  mixture  of  120  pounds  at  80  cents  a  pound  ? 

Sue.  Solve  by  substituting  in  the  formula  obtained  from  the  solution  of 
the  preceding  example. 

20.  If,  for  a  given  distance,  the  rate  is  3  miles  an  hour  over 
the  first  half,  what  must  it  be  over  the  second  half  to  make  the 
average  rate  4  miles  an  hour  ?    5  miles  an  hour  ?    G  miles  an  hour  ? 

21.  The  hind  wheels  and  fore  wheels  of  a  carriage  have  circum- 
ferences 1G  and  14  feet  respectively.  How  far  has  the  carriage 
advanced  when  the  fore  wheels  have  made  51  revolutions  more 
than  the  hind  wheels  ? 

22.  A  train  leaves  A  at  11  a.m.  for  B,  at  the  rate  of  25  miles 
an  hour.  Another  train  leaves  C  at  noon  and  runs  through  A  to 
B  at  the  rate  of  35  miles  an  hour,  arriving  at  B  24  minutes  later 
than  the  first  train.  The  distance  from  C  to  A  being  21  miles, 
find  the  distance  from  A  to  B. 

23.  An  artesian  well  supplies  a  manufactory.  The  water  is 
drawn  out  each  week  day  from  3  a.m.  to  G  p.m.  twice  as  rapidly 
as  it  runs  into  the  well.  If  the  well  contained  2250  gallons  of 
water  on  Monday  morning  and  was  just  emptied  at  G  p.m.  a  week 
from  the  following  Thursday,  how  many  gallons  flowed  into  the 
well  per  hour  ? 

24.  Two  men,  A  and  B,  57  miles  apart,  travel  toward  each 
other,  A  at  the  rate  of  6  miles  an  hour  and  B  at  the  rate  of  5 
miles  an  hour,  B  starting  20  minutes  later  than  A.  How  far  will 
each  have  traveled  when  they  meet  ? 

25.  Find  the  time  between  3  and  4  o'clock  when  the  hands  of 
a  watch  are  opposite  each  other. 

Solution.  Let  x  =  the  time  past  3  ;  then  x  is  also  the  number  of  minute 
spaces  passed  over  by  the  minute  hand,  and  —  is  the  number  passed  over 
by  the  hour  hand,  since  it  moves  fa  as  fast.     Now,  x  in  spaces  is  made  up  of 


152       .  '     HIGHER   ALGEBRA 

three  parts,  viz.,  15  spaces  from  the  XII  mark  to  the  III  mark,  —  from  the 

III  mark  to  where  the  hour  hand  is,  and  30  from  where  the  hour  hand  is  to 
where  the  minute  hand  is  ;  i.e., 

whence  x  =  49^. 


26.  Find  the  time  between  3  and  4  o'clock  when  the  hands  of  a 
watch  are  together,  and  the  tkne  when  they  are  at  tight  angles  to 
each  other. 

27.  Find  the  time  between  4  and  5  o'clock  when  the  minute 
hand  of  a  watch  is  18  minute  spaces  ahead  of  the  hour  hand. 

28.  A  bicyclist,  at  12  miles  an  hour,  went  to  meet  another  who 
started  half  an  hour  later,  at  15  miles  an  hour,  from  a  place  114 
miles  distant.  In  how  many  hours  after  the  second  started  did 
they  meet,  and  how  far  did  each  go  ? 

29.  The  sum  of  the  two  digits  of  a  number  is  7.  If  0  be  sub- 
tracted from  the  number,  the  digits  will  be  interchanged.  What 
is  the  number  ? 

30.  The  left-hand  digit  of  a  number  composed  of  six  digits 
is  1.  If  the  1  be  removed  to  units'  place,  the  other  digits  remain- 
ing in  the  same  order  as  before,  the  new  number  will  be  3  times 
the  original  number.     Find  the  number. 

Sue     Let  x  be  the  number  exclusive  of  the  left-hand  digit. 

31.  Accommodation  trains  leave  A  for  B  at  intervals  of  a  hours,, 
running  m  miles  an  hour.  Express  trains  run  from  B  to  A  at 
the  rate  of  n  miles  an  hour.  At  what  intervals  of  time  does  an 
express  train  meet  the  accommodation  trains  ? 

32.  The  duty  on  a  certain  article  is  reduced  from  $  2.25  per 
hundredweight,  and  in  consequence  of  this  reduction  the  con- 
sumption is  increased  one  half,  but  the  revenue  falls  one  third. 
Find  the  duty  per  hundredweight  after  the  reduction. 

33.  A  man  walking  from  a  town  A  to  another  B  at  the  rate  of 
4  miles  an  hour,  starts  one  hour  before  a  coach  which  goes  12 
miles  an  hour,  and  is  picked  up  by  the  coach.  On  arriving  at  B, 
he  observes  that  his  coach  journey  lasted  2  hours.  Find  the 
distance  from  A  to  B. 


SIMPLE  EQUATIONS  153 

34.  A  hare,  50  of  her  leaps  ahead  of  a  hound,  takes  4  leaps  to 
the  hound's  3 ;  but  2  of  the  hound's  leaps  equal  3  of  the  hare's. 
How  many  leaps  must  the  hound  take  to  catch  the  hare  ? 

35.  A  bicyclist  starts  on  the  road  with  a  message  at  a  rate 
which  will  require  6  hours  for  its  delivery.  At  the  halfway 
point  it  is  taken  by  another  bicyclist  who  rides  3  miles  an  hour 
faster  than  the  first,  and  thus  the  message  is  delivered  half  an 
hour  earlier  than  it  would  have  been  had  the  first  continued. 
Find  the  rate  of  the  first  bicyclist  and  the  whole  distance. 

36.  A  can  do  in  30  days  a  piece  of  work  which  B  can  do  in 
20  days.  A  begins  the  work,  but  after  a  time  B  takes  his  place, 
and  the  whole  work  is  finished  25  days  from  the  beginning.  How 
long  did  A  work  ? 

37.  A  can  do  in  20  days  a  piece  of  work  which  B  can  do  in  30 
days.  A  begins  the  work,  but  later  B  takes  his  place  and  finishes 
it,  working  10  days  longer  than  A.     How  long  did  A  work  ? 

38.  A  man  starts  on  a  bicycle  ride  at  the  rate  of  10  miles  an 
hour,  intending  to  be  back  in  2  hours.  Owing  to  a  breakdown 
he  walks  back  at  the  rate  of  4  miles  an  hour,  and  finds  him- 
self 11  hours  late.     Find  how  far  he  went. 

39.  The  capacity  of  the  second  of  four  casks  is  £  of  the  first, 
the  third  is  f  of  the  second,  the  fourth  is  -^6-  of  the  third,  and 
the  first  holds  15  quarts  more  than  the  third  and  fourth.  How 
many  quarts  does  each  hold  ? 

40.  A  man  invested  part  of  $  2550  in  3%  stocks  and  the  rest 
in  railroad  shares  of  $  25  each,  which  pay  annual  dividends  of 
$  1.00  per  share.  The  stocks  cost  him  $  81  on  a  hundred,  and 
the  railroad  shares  $  24  per  share.  His  income  from  each  source 
is  the  same.  Find  the  number  of  railroad  shares  and  the  amount 
of  each  investment. 

41.  An  express  train  runs  15  miles  an  hour  faster  than  an 
accommodation  train,  and  occupies  T9$  as  much  time  in  running 
120  miles.  The  express  train  loses  at  intermediate  stations  half 
as  much  time  as  does  the  accommodation  train,  and  the  latter 
loses  as  much  time  as  it  would  require  in  running  20  miles. 
Find  the  rate  of  each  train. 


154  HIGHER  ALGEBRA 

SECTION   II  — SIMULTANEOUS   SIMPLE    EQUATIONS    WITH 
TWO    UNKNOWN   QUANTITIES 

254.  Simultaneous  Equations  are  such  as  are  to  be  satisfied  for 
the  same  values  of  the  unknown  quantities.  They  express  differ- 
ent relations  of  the  same  unknown  quantities  and  arise  from 
different  conditions  of  the  same  problem. 

If  we  have  a  single  equation  containing  two  unknown  quanti- 
ties, we  may  assume  any  value  we  please  for  one  and  find  such 
value  for  the  other  as  will  satisfy  the  equation.  Not  so,  however, 
if  the  values  are  to  satisfy  at  the  same  time  two  different  equations, 
such  as  come  from  different  conditions  of  the  same  problem.  If 
the  equations  are  of  the  first  degree,  only  one  set  of  values  will 
satisfy  both.  We  are  now  concerned  with  the  methods  of  finding 
these  values. 

255.  Elimination  is  the  process  of  deducing  from  a  set  of  two 
or  more  simultaneous  equations  containing  as  many  unknown 
quantities  a  new  set  in  which  the  number  of  equations  and  the 
number  of  unknown  quantities  shall  be  diminished  by  at  least  one. 

256.  Three  methods  of  elimination  are  in  common  use,  viz.,  by 
Addition  or  Subtraction,  by  Comparison,  and  by  Substitution. 

257.  The  rules  for  elimination  by  these  different  methods  are 
here  given  and  illustrated.  The  student  should  be  able  to  show, 
first,  why  these  operations  give  true  equations,  and  second,  why 
they  eliminate  one  of  the  unknown  quantities. 

After  eliminating  one  of  the  two  unknown  quantities  from  two 
equations,  the  resulting  equation  is  to  be  solved  for  the  remain- 
ing unknown  quantity.  Usually  the  other  is  found  by  substitut- 
ing this  value  in  the  simpler  of  the  two  given  equations,  and 
solving  the  resulting  equation. 

258.  Prob.      To  eliminate  by  Addition  or  Subtraction. 

Rule.  1st.  If  the  coefficients  of  the  unknown  quantity  to  be  elim- 
inated are  not  numerically  the  same,  make  them  so  by  multiplying 
each  by  that  number  which  will  give  for  the  product  their  1.  c.  m. 


SIMULTANEOUS   SIMPLE  EQUATIONS  155 

2d.  If  the  signs  of  these  coefficients  are  unlike,  add  the  equations; 
if  alike,  subtract  one  equation  from  the  other. 

ILLUSTRATIVE    EXAMPLE 

Sol-  {T  +  ly  =  T 
[4x-  Sy  =  4. 

Multiplying  the  first  by  2  and  the  second  by  3  and  subtracting, 

12  x  +  14^  =  58 

12s-    9y=12 

23  y  =  46 

•••  y  =  2. 

Substituting  this  value  of  y  in  the  second  of  the  given  equations  and 

solving  for  x, 

4x  -6  =  4, 

4z=10, 

x  =  2%. 

Had  we  multiplied  the  first  equation  by  3  and  the  second  by  7  and  added, 
y  would  have  been  eliminated. 

After  finding  the  value  of  one  of  the  unknown  quantities,  the  other  should 
be  found  by  substituting  this  in  the  equation  containing  the  smallest  numbers. 

259.    Prob.     To  eliminate  by  Comparison. 

Rule.     1st.   Find  from  each  equation  the  value  of  the  same  un- 
knoivn  quantity  in  terms  of  the  other  and  known  quantities. 
2d.    Place  these  two  values  equal. 


Solve 


ILLUSTRATIVE    EXAMPLE 

7  x  +    1  y  =  68, 


r  7  x  + 11  y  =  m 

\9x  -4y  =  33. 


/>Q    11.. 

From  the  first,  x  = = 

33  +  4y 


From  the  second,  x 

Hence, 


9 
68-lly     33 +  4  y 


7  9 

612-    99?/ =  231  + 28 y, 
-  127  y  =-381, 
y  =  3. 


156  HIGHER  ALGEBRA 

Substituting  this  value  of  y  in  the  second  of  the  given  equations  and 
solving  for  as, 

9x-  12  =  33, 

9x  =  45, 

x  =  5. 

260.    Prob.      To  eliminate  by  Substitution. 

Rule.  1st.  Find  from  one  of  the  equations  the  value  of  one  of 
the  unknown  quantities  in  terms  of  the  other  and  known  quantities. 

2d.  Substitute  this  value  for  the  same  unknown  quantity  in  the 
other  equation. 

ILLUSTRATIVE    EXAMPLE 

lx-2y  =  l,  (1) 

3x  +  5y  =  59.  (2) 

From(l),  y^^f1'  & 

Substituting  this  value  of  y  in  (2)  and  solving  for  x, 

7ic  —  1 


Solve 


3s  +  5[ 


59, 

2       / 

6z  +  35z-5  =  118, 

41  x  =  123, 

x  =  S. 
Substituting  this  value  of  x  in  (3), 

SOME  PRACTICAL   SUGGESTIONS 

261.  1.  When  using  the  method  by  Addition  or  Subtraction,  if 
one  of  the  coefficients  of  one  of  the  unknown  quantities  is  itself 
the  1.  c.  m.  of  these  coefficients,  eliminate  that  unknown  quan- 
tity, inasmuch  as  this  requires  that  only  one  of  the  equations  be 
multiplied.  Otherwise,  eliminate  the  unknown  quantity  whose 
coefficients  require  the  smallest  multipliers  to  make  them  the 
same. 

2.  After  finding  the  value  of  one  of  the  unknown  quantities, 
the  other  should  be  found  by  substituting  this  in  the  equation 
containing  the  smallest  numbers,  or,  in  case  of  literal  equations, 
the  simplest  coefficients. 


SIMULTANEOUS   SIMPLE  EQUATIONS 


157 


3.  When  either  of  the  equations,  after  reduction,  contains  in 
more  than  one  term  the  unknown  quantity  to  be  eliminated,  it  is 
not  expedient  to  use  the  method  by  Addition  or  Subtraction,  inas- 
much as  no  monomial  multiplier  will  make  the  coefficients  alike. 

4.  When  the  unknown  quantity  to  be  eliminated  occurs  in  a 
monomial  denominator  in  both  equations,  do  not  clear  of  frac- 
tions before  eliminating.  Such  equations  may  thus  be  solved  by 
the  methods  for  simple  equations,  although  they  are  of  the  second 
degree. 

EXAMPLES  LXXI 


Solve  the  following,  using  all  the  different  methods  of  elimina- 
tion : 


i'3x  —  4y  =  2, 
[  7  x  -  9  y  =*  7. 

11  x  -  7  y  =  37, 
8x  +  9*/ =  41. 


2  x  +  7  y  =  41, 

3  x  +  4  y  =  42. 


|5a> 
\l  x 


5  x  -  3  y  =  4, 

12  7/  =  -10. 


5    j3z-4*/  =  18, 
{3x  +  2y  =  0. 

f  15  a  +  7  y  =  29, 
1  9  x  +  15  y  =  39. 


\2x-5y  =  -21, 
13  x  -  4  y  =  120. 

f  22  x  -f  15  y  =  9, 
18  a  +  25  y  =  71. 


18  x  -  20  y  =  44, 
17  x  -  15  y  =  26. 


10. 


17  x  -  69  2/  =  -  103, 
14  «  -  13  y  =  -  41. 


11. 


2x  ,  3y         7 
T+4       -2' 


x     2y 
4       5 


11 

2' 


12 


a  -  1     y  -  1 
5  7 


0, 


,2x-3      22/  + 13 


=  0. 


13. 


10  x 


Sy- 


^  =  -17. 


14. 


6  +  x-y=  7} 
1  —  x  —  y  4 
2a;  +  3#  =  -l. 


158 


HIGHER  ALGEBRA 


15. 


17. 


19 


21. 


23. 


25. 


2x 
3 


5y 
12 


Sx 


f 

4 


23 

2 


=  2, 


I 

.i 


aj  +  y     o 
ax  +  by  =  c, 
a'x  +  b'y  =  c'. 

(Hc)a;+(6-c)?/=2a&, 
(a -f- c)  a  —  (a  —  c)y=2  ac. 

10     9 

SB       2/ 
8_15 

x      y 


=  4, 

9 

2" 


£-1  =  2. 

ax      by 


a       b 
bx     ay 


a  +  b, 


x      y 


20. 


22. 


24. 


26. 


16. 


18.    \ 


x  —  3  y     y  —  3  g 

2  2 

30 


(  ax  —  by  =  2  a6, 

I  2  6a;  +  2  ay  =  3  62  -  a2. 
(a-j-6)a;—  (a— b)y=3ab, 
(a—b)  x— (a-\-b)y =ab. 

_5__7  =  29 
3a;     y       9' 

3  ,  JL  =  _2. 

u     4?/         8 


6a; 


«'/ 


=  0, 


aa;      by      a2 


a  +  b 


x 


ay 


ab 


a;  ?/ 


b 

— , 

a 
g2+3«6 
a  +  b 


PROBLEMS  LEADING  TO   SIMPLE   EQUATIONS   WITH  TWO 
UNKNOWN   QUANTITIES 

EXAMPLES  LXXII 

1.  The  daily  pay  of  5  men  and  4  boys  is  $  19,  and  the  daily 
pay  of  2  men  exceeds  that  of  3. boys  by  $  3.  What  is  the  daily 
pay  of  each  man  and  boy  ? 

Solution.     Let  x  —  the  daily  pay  of  each  man, 

and  y  =  the  daily  pay  of  each  boy  ; 

then  6  x  +  4  y  =  19, 

and  2  x  -  -  3  y  =  3. 

From  these  equations  we  find  x  =  3  and  y  =  1. 


SIMULTANEOUS   SIMPLE  EQUATIONS 


159 


Solution. 
and 

Then 
and 


Let 


2.  The  sum  of  the  two  digits  of  a  number  is  13,  and  the  num- 
ber diminished  by  that  formed  by  reversing  the  digits  is  27. 
Find  the  number. 

x  =  the  digit  in  tens'  place, 
y  =  the  digit  in  units'  place. 
10  x  +  y  =  the  number, 

10  y  +  x  =  the  number  formed  by  reversing  the 
digits. 
Hence,  x  +  y  =  13, 

and  I0x  +  y  -(\0y  +  x)  =  2~. 

From  these  equations  we  find  x  =  8  and  y  =  5. 
Hence  the  number  is  10  x  8  +  5  =  85. 

3.  An  income  of  $  120  a  year  is  derived  from  a  sum  of  money 
invested,  partly  in  3J  per  cent  stock  and  partly  in  4  per  cent 
stock.  If  the  stock  be  sold  when  the  first  is  at  108  and  the  second 
at  120,  the  sum  realized  will  be  $  3672.     Find  each  investment. 

Solution. 
and 
Then 


Let 


income  from  the  first, 


x  =  the  3£  per  cent  stock, 
y  =  the  4  per  cent  stock. 

100 

4y 

100 

108  a; 


100 


income  from  the  second, 
selling  price  of  the  first, 


and 


Hence, 


and 


y~  =  selling  price  of  the  second. 

100 

^  +  il=120, 
100      100 

M£  +  !20j/  =  3672. 


100         100 
From  these  equations  we  find  x  =  2400  and  y  =  900. 

4.  One  kind  of  wine  is  worth  72  cents  a  quart,  and  another 
40  cents.  How  much  of  each  must  be  put  into  a  mixture  of 
50  quarts  that  shall  be  worth  60  cents  a  quart? 

5.  If  6  pounds  of  sugar  and  10  pounds  of  tea  cost  $  6.30,  and, 
at  the  same  price,  10  pounds  of  sugar  and  6  pounds  of  tea  cost 
$  4.10,  what  is  the  price  of  each  per  pound  ? 


160  HIGHER   ALGEBRA 

6.  A  farmer  bought  120  acres  of  land  for  $13,200,  paying 
$  80  an  acre  for  part  of  it,  and  $  120  an  acre  for  the  remainder. 
Find  the  number  of  acres  in  each  part. 

7.  If  5  be  added  to  the  numerator  of  a  certain  fraction,  its 
value  becomes  j ;  and  if  5  be  subtracted  from  its  denominator,  its 
value  becomes  f .     Find  the  fraction. 

8.  A  crew  that  can  row  12  miles  an  hour  downstream  finds 
that  it  takes  twice  as  long  to  row  a  given  distance  upstream. 
Find  the  rate  of  the  current  and  the  rate  of  the  crew  in  still 
water. 

9.  A  certain  number  is  equal  to  4  times  the  sum  of  its  two 
digits,  and  if  18  be  added  to  it,  the  digits  will  be  reversed.  What 
is  the  number  ? 

10.  A  and  B  can  do  a  piece  of  work  in  9  hours.  After  working 
together  7  hours,  B  finishes  the  work  in  5  hours  more.  In  how 
many  hours  could  each  do  the  work  ? 

11.  In  an  informal  ballot  a  resolution  was  adopted  by  a 
majority  of  10  votes ;  but  in  the  formal  ballot  one  fourth  of  those 
who  had  before  voted  for  it  voted  against  it,  and  the  resolution 
was  lost  by  a  majority  of  6  votes.  How  many  voted  each  way  in 
the  formal  ballot  ? 

12.  When  a  is  added  to  the  greater  of  two  numbers,  the  sum  is 
m  times  the  less ;  but  when  b  is  added  to  the  less,  the  sum  is  n 
times  the  greater.     Find  the  numbers. 

13.  When  4  is  added  to  the  greater  of  two  numbers,  the  sum  is 
3^  times  the  less ;  but  when  8  is  added  to  the  less,  the  sum  is  ^ 
the  greater.  Find  the  numbers  by  substituting  in  the  results  of 
the  preceding  example. 

14.  If  two  trains,  100  miles  apart,  approach  each  other,  they 
will  meet  in  2  hours ;  but  if  they  run  in  the  same  direction,  the 
slower  train  leading,  they  will  be  together  in  10  hours.  What  is 
the  rate  of  each  train  ? 

15.  A  father  has  two  sons,  one  4  years  older  than  the  other. 
Six  years  ago  the  father's  age  was  6  times  the  joint  ages  of  his 
sons,  while  2  years  hence  his  age  will  be  twice  the  joint  ages  of 
his  sons.     Find  the  age  of  each. 


SIMULTANEOUS   SIMPLE  EQUATIONS  161 

16.  A  and  B  can  together  do  a  certain  work  in  30  days ;  at  the 
end  of  18  days,  however,  B  is  called  off,  and  A  finishes  it  alone  in 
20  days  more.     Find  the  time  in  which  each  could  do  the  work. 

17.  A  man  rows  30  miles  down  a  river  and  then  back  to  the 
starting  place,  his  whole  time  being  12  hours.  He  finds  that  he 
can  row  5  miles  with  the  stream  in  the  same  time  as  3  against  it. 
Find  his  time  down  and  up  respectively. 

18.  In  an  alloy  of  silver  and  copper  —  of  the  whole  -\-p  ounces 

-.  m 

was   silver,  and   -  of   the  whole  —  q  ounces  was  copper.     How 
n 

many  ounces  were  there  of  each  ? 

19.  A  man  invests  $  5100,  partly  in  3|  per  cent  stock  at  90  and 
partly  in  4  per  cent  stock  at  120,  and  receives  from  the  two 
investments  $  185  a  year.  How  many  shares  of  each  stock  does 
he  buy  ? 

20.  A  man  sculls  in  1  hr.  20  min.  a  certain  distance  down  a 
stream  which  runs  at  the  rate  of  4  miles  an  hour.  In  returning 
it  takes  him  4  hr.  15  min.  to  reach  a  point  3  miles  below  his  start- 
ing place.  How  far  did  he  scull  down  the  stream,  and  at  what 
rate  could  he  scull  in  still  water  ? 

21.  A  tank  is  supplied  by  two  pipes.  If  the  first  be  opened  6 
minutes  and  the  second  7  minutes,  the  tank  will  be  filled ;  or  if 
the  first  be  opened  3  minutes  and  the  second  12  minutes,  the  tank 
will  be  filled.     In  what  time  will  each  pipe  fill  the  tank  ? 

22.  A  cistern  is  supplied  by  three  pipes,  two  of  which  are  of  the, 
same  size.  When  they  are  all  open,  T5^  of  the  cistern  is  filled  in 
4  hours ;  but  if  one  of  the  equal  pipes  be  closed,  J  of  the  cistern 
is  filled  in  10|  hours.  In  how  many  hours  would  each  pipe  fill 
the  cistern  ? 

23.  An  alloy  of  tin  and  lead,  weighing  40  pounds,  loses  4 
pounds  when  immersed  in  water.  It  is  found  that  10  pounds  of 
tin  lose  1.375  pounds  when  immersed  in  water,  and  5  pounds  of 
lead  lose  .375  pounds.  How  many  pounds  of  each  metal  are  in  the 
alloy  ? 

downey's  alg.  — 11 


162  BIGHER  ALGEBRA 

24.  A  man  invests  $  17,200  in  3  per  cent  bonds  at  90  and  5  per 
cent  bonds  at  108,  and  his  incomes  from  the  two  investments  are 
the  same.     Find  the  amount  of  each  investment. 

25.  A  dairyman  mixed  with  100  quarts  of  morning's  milk  the 
night's  milk,  from  which  the  cream  had  been  skimmed,  selling 
the  mixture  for  5  cents  per  quart  and  the  cream  for  20  cents  per 
quart.  By  this  fraudulent  means  he  realized  $  1.80  more  than 
he  would  had  he  sold  at  5  cents  per  quart  both  the  morning's 
milk  and  the  night's  milk  without  skimming.  How  much 
skimmed  milk  and  how  much  cream  did  he  sell? 

26.  B  has  $  1000  more  capital  than  A,  invests  it  at  one  per 
cent  more,  and  receives  $  80  more  income.  C  has  $  500  more 
capital  than  B,  invests  it  at  one  per  cent  more,  and  receives  $  70 
more  income.     Find  A's  capital  and  rate. 

27.  Two  men,  A  and  B,  are  employed  on  a  piece  of  work.  A 
works  1|  days  by  himself,  when  B  joins  him,  and  they  complete 
T83  of  the  work  2\  days  later ;  they  then  find  that  2  days  more 
will  be  required  for  them  to  finish  the  work.  In  how  many  days 
could  each  do  the  work  ? 

28.  If  a  pieces  of  one  kind  of  money  make  a  dollar,  and  b  pieces 
of  another  kind  make  a  dollar,  how  many  pieces  of  each  kind 
must  be  taken  to  have  c  pieces  in  a  dollar  ? 

29.  The  distance  from  A  to  B  along  a  railway  is  70  miles,  the 
first  10  miles  being  level,  the  next  35  sloping  upward,  and  the 
rest  level.  A  train  starting  from  A  runs  half  the  distance  in  62 
minutes,  and  the  whole  distance  in  1  hour  52  minutes.  Find  the 
rates  of  the  train  on  the  level  ground  and  on  the  up  grade. 

30.  Two  cyclists,  A  and  B,  ride  a  race,  the  course  being  from 
P  to  Q,  a  distance  of  12  miles,  and  back.  A  gives  B  a  start  of  15 
minutes,  and  meets  him  on  his  return  journey  1680  yards  from  Q, 
afterward  winning  the  race  by  1  minute.  Find  the  rate  of  each 
cyclist,  assuming  it  uniform. 

31.  Two  men  received  $  96  for  a  piece  of  work  which  they  could 
do  together  in  30  days.  When  half  the  work  was  done,  one  of 
them  stopped  8  days  and  the  other  4  days.  They  completed  the 
work  in  35^  days  from  the  beginning.  How  long  would  each 
require  to  do  the  work,  and  how  many  dollars  should  each  receive  ? 


SIMULTANEOUS   SIMPLE  EQUATIONS  163 

32.  A  man  in  a  rowboat  is  at  a  distance  a  from  two  barges  at 
the  instant  when  they  are  passing  each  other,  one  coming  toward 
him  and  the  other  going  away  from  him,  the  two  having  the  same 
rate.  Show  that  if  b  and  b'  are  the  distances  he  rows  before 
meeting  one  and  overtaking  the  other, 

?  =  !+!. 

a      b      b' 

33.  A  and  B  formed  a  partnership.  A  invested  $  20,000  of  his 
own  money  and  $  5000  which  he  borrowed ;  B  invested  $  22,000  of 
his  own  money  and  $  8000  which  he  borrowed  at  the  same  rate  of 
interest  as  was  paid  by  A.  At  the  end  of  a  year  A's  share  of  the 
profits  was  $  1750  more  than  the  interest  on  his  $  5000,  and  B's 
was  $2000  more  than  the  interest  on  his  $8000.  What  rate  of 
interest  did  they  pay,  and  what  rate  per  cent  did  they  realize  on 
their  investment  ? 

34.  An  ingot  of  metal  which  weighs  n  pounds  loses  p  pounds 
when  weighed  in  water.  This  ingot  is  itself  composed  of  two 
other  metals,  which  we  may  call  A  and  B.  Now  n  pounds  of  A 
lose  q  pounds  when  weighed  in  water,  and  n  pounds  of  B  lose  r 
pounds  when  weighed  in  water.  How  much  of  each  metal  does 
the  original  ingot  contain  ? 

35.  A  body  moves  with  uniform  velocity  from  A  to  B,  323 
feet,  and,  without  stopping,  returns.  A  second  body  leaves  B  13 
seconds  after  the  first  leaves  A  and  moves  toward  A  with  uniform 
velocity.  The  first  body  meets  the  second  10  seconds  after  the 
latter  starts,  and,  in  returning  to  A,  overtakes  the  second  body  45 
seconds  after  the  latter  starts.     Find  the  velocity  of  each  body. 

SECTION  III  — SIMULTANEOUS  SIMPLE   EQUATIONS  WITH 
SEVERAL   UNKNOWN   QUANTITIES 

262.  Prob.  To  solve  several  simultaneous  simple  equations  with 
as  many  unknown  quantities. 

Rule.  1st.  Eliminate  the  same  unknown  quantity  from  different 
pairs  of  the  given  equations,  thus  forming  a  set  of  equations  inde- 
pendent of  this  unknown  quantity,  and  one  less  in  number  than  the 
given  equations. 


164 


HIGHER  ALGEBRA 


2d.  From  these,  in  like  manner,  eliminate  another  unknown  quan- 
tity, and  so  continue  till  an  equation  with  but  one  unknown  quantity 
is  found. 

3d.  Find  the  value  of  this  unknown  quantity  and  substitute  it  in 
one  of  the  two  equations  of  the  next  preceding  set.  Solve  this  equa- 
tion and  substitute  the  two  values  now  found  in  one  of  the  three  equa- 
tions of  the  next  preceding  set.  Continue  this  jwocess  till  all  the 
unknown  quayitities  are  determined. 

263.  Sch.  1.  It  is  usually  best  to  combine  the  equation  hav- 
ing the  smallest  coefficients  with  each  of  the  other  equations  of 
the  same  set. 

264.  Sch.  2.  If  any  equation  of  any  set  does  not  contain  the 
unknown  quantity  we  are  eliminating,  this  equation  is  written 
unchanged  in  the  next  set. 


l. 


EXAMPLES  LXXIII 
Solve  the  following : 

(3x  +  4y  —  z  =  8, 
x  +  2y-4;Z  =  -7, 
2x-5y-7z  =  -29. 

2x  +  3y-4z  =  8, 
3x-±y  +  2z  =  3, 
4:X  —  2y—'3z  =  5. 


7. 


{2x-3y  =  4, 
4x-3z=2, 
4?/  +  2z=-3. 

w  -\-3x  —  y  —  z  =  7, 
2w  —  2x  +  y  +  3z  =  8, 
3w  —  x  +  y  —  4«  =  8, 
4w  +  a;  —  y  —  2z  =  7. 


4. 


2a?-f3y  +  3  =  17, 
2  x  4-  2  y  +  z  ===  14, 

x  +  3y  +  2z  =  Vd. 

12x  —  ky  +  z  =  3, 
x—y  —  2z  —  —  l, 
[5x-2y  =  0. 

y  +  z 


x  + 


85, 


6.  ^  +  ^  =  85, 


z  + 


x  +  y  _ 


=  85. 


2x  +  3y  +  z  =  23, 
2u  +  3x  +  y  =  25, 
u  +  x  +  3z  =  24, 
[3u  +  2y  +  2z  =  36. 


SIMULTANEOUS   SIMPLE  EQUATIONS 


165 


u-2x  =  -13, 
x-3y=13, 
y-±z  =  5, 


10. 


ou 


23. 


u  +  5x  —  7  y  +  z  =  2, 
2u  +  7x-3y-3z  =  2y 
4x-2y  =  2, 
ix  +  5y  —  2z  =  2. 


Although  the  following  are  not  of  the  first  degree  with  reference  to  x,  y, 
and  z,  they  are  of  the  first  degree  with  reference  to  the  reciprocals  of  these 


We  may,  therefore,  regard    ,  -,  and  -  as  the  unknown  quanti- 
se y  z 


quantities 

ties  and  solve  accordingly,  i.e.,  eliminate  without  first  clearing  of  fractions, 


il- 


ia. 


15. 


fl    1 

1  _ 

z 

3, 

1.-14 

x      y 

1 

z  ~ 

•2, 

M- 

lx      y 

1_ 

z  ~ 

:1. 

[1+1- 

x      y 

2 
=  3' 

1      1_ 

x     z~ 

3 
4' 

[y     z 

5 
=  6' 

a     b 
x     y~ 

=  1, 

6  +  °- 

y    z 

=  1, 

c     a 

.  z     x~ 

:1. 

12. 


14. 


16. 


f2 

X 

1 

+  y- 

2; 

2 

y 

3_ 

z  ~ 

-2, 

l 

.  x 

1 

+  ~z  = 

4 
3* 

1 

X 

1 

+-y+ 

z 

2 

X 

3 

z 

5 
.  x 

* 

5  =  4. 

z 

1 

X 

+A 

9 

"5' 

1 

y 

+& 

5 

"3' 

l 

.z 

4sc 

7 
4* 

In  solving  the  next  two  examples,  divide  each  equation  by  the  unknown 
quantities  in  the  second  member,  and  then  eliminate  fractional  terms. 


17. 


y  +  z  =  2yz, 
x  +  z  =  3  xz, 
x  +  y  =  4xy. 


yz+k  xz—3  xy=2xyz, 

\  3yz—2xz-\-6xy=4:xyz, 

5  yz—6  xz—3  xy=xyz. 


166 


HIGHER  ALGEBRA 


In  solving  the  next  two  examples,  let  xyz  constitute  one  member  of  each 
equation,  divide  each  equation  by  xyz,  and  then  eliminate  fractional  terms. 

'  19.   xyz  =  2  (xz  -f-  xy  —  yz)  =  3  (xz  -f-  yz  —  xy)  =  4  (xy  -\-yz  —  xz). 

20.   xyz  =  a(yz  —  xz  —  xy)  =  b  (xz  —  xy  —  yz)  =  c  (xy  —  yz  —  #z). 

When  one  of  the  unknown  quantities  is  wanting  in  each  of  the  given  equa- 
tions, and  all  the  coefficients  (after  simplifying  the  equations)  are  unity,  the 
work  may  be  made  very  short  by  adding  all  the  equations,  dividing  by  the 
common  coefficient,  and  then  subtracting  from  the  resulting  equation  each  of 
the  given  equations  in  turn.  Likewise,  when  all  of  the  unknown  quantities 
are  present  in  each  equation,  and  all  of  the  coefficients  (after  simplifying  the 
equations)  except  one  are  unity,  and  this  one  coefficient  is  repeated  with 
successive  unknown  quantities  in  successive  equations,  the  work  may  be 
made  very  short  by  adding  all  the  equations,  dividing  by  the  common  coeffi- 
cient, and  then  subtracting  the  resulting  equation  from  each  of  the  given 
equations  in  turn.     Solve  in  this  manner  the  following  : 


(x  +  y  =  9, 
21.  I  x  4-  *  =  10, 
I  y  +  z  =  11. 


22. 


23. 


25. 


27. 


fw  +  x  +  y  =  3, 
w  -\-  x  +  z  =  —  4, 
iv  +  y  +  z  =  2, 
x  +  y  +  z=-l. 

v  +  w  +  x  -j-  y  =  10, 
v  +  iv  -f  x  -f-  z  =  11, 
v  +  w  +  y  -f  z  =  12, 
v  +  x  +  y  +  z  =  13, 
w  +  x  +  y  -f  z  =  14. 

3w  +  x  +  y  +  z  =  l4:, 

w  +  3x  +  y  +  z  =  l2, 

w  4-  x '  +  3y  +  z  =  16, 

[w  +  x  +  y  +  3z  =  18. 


24. 


26. 


28. 


w  +  x  +  y  =  l, 
w  -f  it-  -f  z  =  6, 
iu  +  ?/  +  z  =  9, 
x  +  y  +  z  =  8. 

w  +  x  +  y  =  Q, 
iv  -f  x  -(-  z  =  6, 
w  +  ?/  +  z  =  —  1, 
#  -f-  ?/  +  z  =  —  2. 

f  u  -f-  v  -f- i<;  +  x  -f  ?/  =  3, 
u +  v  +  w  -f-  x  +  z  =  8, 
w  +  v-j-w-f2/-fz  =  4, 
**  +  v  -f-  x  +  y  +  k  «  9, 
m  -f-  to  +  x  -f-  ?/  +  z  =  6, 

lv  +  w4-aj-r-y4-2!  =  5. 

(w-f-#  +  ?/-|-4z  =  —  9, 
w-f-#-f-4?/-|-z  =  3, 
w  +  4#-f-y-f-z  =  —  6, 
4w  +  x-f-?/-f-z  =  12. 


SIMULTANEOUS  SIMPLE  EQUATIONS 


167 


29. 


2w+2x+2y+3z=39, 
2tv-{-2x-\-3y-\-2z=37, 
2w+3x+2y+2z=35, 
3w+2x+2y+2z=33. 


'  v  +  w  -f-  x  +  y  +  2  z  =  52, 

v  +  w  +  g  +  2  ?/  +  z  =  50, 

31.  {t>  +  to+2a?+y+s=48, 

v  4-  2  w  +  a;  +  y  +  z  =  46, 
12v  +  M7  +  aj  +  y  +  z  =  44. 


30. 


fw4-5a;4-2/4-z  =  22, 
iu  -f  x  +  y  4-  5  z  =  26, 
5w  +  x-f?/  +  z  =  18, 

.  ic  4-  a;  4-  5  y  4-  2  =  —  18. 


'  v  4- w  4- 6  #4- 1/4-2= 40, 
6v-M«4-a;4-2/4-z=36, 
32.   j  v+w+ajH-6y+2=lj 

v+w-f-o;+?/-|-62;=31. 


PROBLEMS  LEADING  TO   SIMPLE    EQUATIONS   WITH   SEVERAL 
UNKNOWN   QUANTITIES 

EXAMPLES  LXXIV 

1.  The  sum  of  the  three  digits  of  a  number  is  12 ;  the  digit  in 
the  tens'  place  is  \  the  sum  of  the  other  two,  and  the  number  ex- 
pressed by  the  two  left-hand  digits  is  7  times  the  digit  in  units' 
place.     Find  the  number. 

2.  For  $  8  I  can  buy  2  pounds  of  tea,  10  pounds  of  coffee,  and 
20  pounds  of  sugar ;  or  2  pounds  of  tea,  5  pounds  of  coffee,  and 
30  pounds  of  sugar;  or  3  pounds  of  tea,  5  pounds  of  coffee, 
and  10  pounds  of  sugar.    .  What  are  the  prices  ? 

3.  Three  cities,  A,  B,  and  C,  not  in  the  same  straight  line,  are 
connected  by  straight  roads.  The  distance  from  A  to  C  by  way 
of  B  is  82  miles,  from  B  to  A  by  way  of  C  is  97  miles,  and 
from  G  to  B  by  way  of  A  is  89  miles.  Find  the  distances  be- 
tween the  cities. 

Sug.  In  solving  the  equations,  proceed  as  directed  for  examples  21  to  2G 
of  the  preceding  set. 

4.  The  total  capacity  of  3  casks  is  1440  quarts.  Two  of 
them  are  full,  and  one  is  empty.  To  fill  the  empty  one  requires 
the  contents  of  the  first  and  ^  the  contents  of  the  second,  or  the 
contents  of  the  second  and  ^  the  contents  of  the  first.  Find  the 
capacity  of  each  cask. 


168  HIGHER  ALGEBRA 

5.  A  farm  was  rented  for  2  years  for  a  fixed  money  payment 
and  400  bushels  of  wheat  and  barley.  The  first  year  wheat  was 
70  cents  per  bushel  and  barley  50  cents,  and  the  entire  rent  was 
$  1000.  The  second  year  wheat  was  50  cents  per  bushel  and 
barley  45  cents,  and  the  entire  rent  was  $  950.  Find  the  amounts 
of  money,  wheat,  and  barley  paid  as  rent  each  year. 

6.  When  3  partners  began  business,  A  had  $  2000  more  than 
twice  as  much  capital  as  B,  and  C  had  $  500  less  than  A  and  B 
together.  The  first  year  A  gained  as  much  as  B's  capital,  B 
gained  as  much  as  A's  capital,  and  C  gained  as  much  as  A's  and 
B's  capital  together,  whereupon  each  had  the  same  sum.  Find 
how  much  each  had  at  first,  and  interpret  the  results. 

7.  The  capacity  of  3  casks  is  344  gallons,  and  all  are  full. 
Fifty  gallons  are  used  from  the  first;  then  i  of  what  is  in  the 
second  is  poured  into  the  first,  and  £  of  what  is  in  the  third  is 
poured  into  the  second.  After  these  changes,  the  first  contains 
10  gallons  more  than  the  second,  and  the  second  10  gallons  more 
than  the  third.     Find  the  capacity  of  each  cask. 

8.  In  walking  along  a  street  on  which  electric  cars  are  running 
at  equal  intervals  from  both  ends,  I  observe  that  I  am  overtaken 
by  a  car  every  12  minutes,  and  that  I  meet  one  every  4  minutes. 
What  are  the  relative  rates  of  myself  and  the  cars,  and  at  what 
intervals  of  time  do  the  cars  start  ? 

9.  Four  towns,  A,  B,  C,  and  D,  connected  by  rail,  are  at  the 
vertices  of  a  quadrilateral.  A  commercial  traveler,  in  making 
the  rounds  of  these  towns,  observes  that  in  going  from  A  by  way 
of  B  and  C  to  D,  the  fare,  at  2  cents  per  mile,  is  $  1.22,  and  in 
going  from  A  by  way  of  D  and  C  to  B,  it  is  $  1.10 ;  while  from  A 
by  way  of  B  to  C,  it  is  the  same  as  from  A  by  way  of  D  to  C,  and 
from  B  by  way  of  A  to  D,  it  is  40  cents  less  than  from  B  by  way 
of  C  to  D.     Find  the  distances. 

10.  A,  B,  C,  and  D  engage  to  do  a  certain  work.  A  and  B  can 
do  it  in  12  days,  A  and  D  in  15  days,  and  D  and  C  in  18  days. 
B  and  C  begin  the  work  together,  after  3  days  are  joined  by  A, 
and  after  4  days  more  by  D.  Then  all  working  together  they 
finish  it  in  2  days  more.  How  long  would  each  have  required  to 
do  the  entire  work  ? 


CHAPTER    XII 
INEQUALITIES 

265.  An  Inequality  is  an  assertion  by  means  of  a  mathematical 
symbol  that  two  expressions  have  different  values.  This  symbol 
is  the  Sign  of  Inequality  >  or  <,  read  "greater  than"  or  "less 
than,"  according  as  the  opening  is  to  the  left  or  right. 

266.  The  Members  of  an  Inequality  are  the  expressions  con- 
nected by  the  sign  of  inequality,  the  one  on  the  left  being  called 
the  First  Member,  and  the  one  on  the  right  the  Second  Member. 

267.  Of  two  positive  quantities,  the  greater  is  the  one  which  is 
numerically  the  greater;  while  of  two  negative  quantities  the 
greater  is  the  one  which  is  numerically  the  less.  A  negative 
quantity  is  less  than  a  positive  quantity,  regardless  of  their 
numerical  values.  In  general,  when  a  —  b  is  positive,  a>b,  and 
when  a  —  b  is  negative,  a<b. 

268.  Two  inequalities  exist  in  the  same  sense  or  exist  in  an 
opposite  sense,  according  as  their  signs  of  inequality  are  alike  or 
opposite. 

269.  Theorem.  An  inequality  will  continue  to  exist  in  the  same 
sense, 

1st.  By  adding  equals  to  both  members,  or  subtracting  equals 
from  both  members. 

c2d.  By  multiplying  or  dividing  both  members  by  equal  positive 
quantities. 

3d.   By  raising  both  members  to  the  same  odd  power. 

4th.  By  raising  both  members  to  the  same  even  power,  if  both 
members  are  positive. 

5th.  By  extracting  the  same  root  of  both  members,  if  when  the 
index  is  even,  only  the  positive  roots  be  compared. 

169 


170  HIGHER  ALGEBRA 

Dem.  Each  of  the  operations  of  the  1st  and  2d  increases  or 
decreases  both  members  alike,  either  by  the  same  amount  or 
in  the  same  ratio,  and  does  not  change  the  signs  of  the  members. 
Therefore,  that  member  which  was  greater  before  any  of  the 
operations  named  is  greater  after. 

Since  the  same  power  or  root  of  the  numerically  greater  of  two 
quantities  is  numerically  the  greater,  and  the  operations  of  the  3d, 
4th,  and  5th  do  not  change  the  signs  of  the  members,  that  member 
which  was  greater  before  any  of  the  operations  named  is  greater 
after. 

270.  Cor.  i.  An  inequality  may  be  cleared  of  fractions  without 
changing  the  sense  in  which  it  exists. 

This  is  simply  multiplying  both  members  by  the  same  positive 
quantity. 

271.  .  Cor.  2.  A  term,  by  changing  its  sign,  may  be  transposed  from 
one  member  of  an  inequality  to  the  other  without  changing  the  sense 
in  ivhich  it  exists. 

This  is  simply  subtracting  the  same  quantity  from,  or  adding 
the  same  quantity  to,  both  members. 

272.  Theorem.  An  inequality  will  be  made  to  exist  in  an  opposite 
sense, 

1st.    By  changing  the  signs  of  both  members. 

2d.  By  multiplying  or  dividing  both  members  by  the  same  negative 
quantity. 

3d.  By  raising  both  members  to  the  same  even  power,  if  both 
members  are  negative. 

4th.   By  comparing  the  negative  even  roots. 

Dem.  The  members  are  affected  numerically  as  in  the  pre- 
ceding demonstration,  but  the  signs  are  changed.  Therefore,  that 
member  which  was  greater  before  any  of  the  operations  named  is 
less  after. 

273.  Theorem.  If  two  or  more  inequalities  exist  in  the  same 
sense, 

1st.    The  sums  will  exist  in  the  same  sense  as  the  given  inequalities. 
2d.    The  products  will  exist  in  the  same  sense  as  the  given  inequali- 
ties, if  their  members  are  all  positive. 


INEQUALITIES  171 

Dem.  In  the  1st  case  the  greater  member  of  any  one  of  the 
inequalities  is  increased  by  a  greater  amount  or  decreased  by  a 
less  amount  than  the  less  member ;  and  in  the  2d  case  the  greater 
member  is  increased  in  a  greater  ratio  or  diminished  in  a  less  ratio 
than  the  less  member. 

274.  Caution.  The  student  is  cautioned  against  taking  the 
differences  or  the  quotients  of  two  inequalities  that  exist  in  the 
same  sense,  inasmuch  as  the  differences  and  quotients  will  exist  in 
the  same  sense  as,  or  the  opposite  sense  from,  the  given  inequali- 
ties, according  to  the  order  of  subtracting  or  dividing,  or  will  give, 
with  certain  relations,  an  equation. 

275.  Note.  In  determining  which  of  two  expressions  is  the 
greater  it  is  sometimes  best  to  place  both  signs,  *S,  read  "greater 
or  less,"  between  them  and  then  operate  on  the  inequality  in  such 
ways  as  not  to  change  the  sense,  until  it  becomes  apparent  which 
is  the  greater  member.  This  will  show  which  of  the  two  signs  to 
use  in  the  original  inequality. 

EXAMPLES    LXXV 

It  is  understood  that  the  letters  of  each  example  are  positive 
and  unequal. 

1.  Which  is  greater,  the  sum  of  the  squares  of  any  two  quanti- 
ties or  twice  their  product  ? 


Solution.     We  have 

a2  +  b2>2ab. 

Transposing, 

a2_2a&-f  &2>o, 

(a-b)2>0. 

But  (a  —  b)2  is  necessarily  plus  and,  therefore,  greater  than  0.  Hence, 
since  the  operations  leave  the  inequality  in  the  same  sense  as  the  first, 

•    a2  +  62  >  2  ab. 

2.  Which  is  greater,  half  the  sum  of  two  quantities  (their 
arithmetical  mean)  or  the  square  root  of  their  product  (their 
geometrical  mean)  ? 


172  HIGHER  ALGEBRA 

Solution.     We  have  ^-±-^  >  Vab. 

2      < 

Squaring,  etc.,  a2  +  2  ab  +  62  J  4  «&, 

a2-2a&  +  ft2>o, 

(«-6)2<0. 

But  (a  — 6)2>  0,  because  the  square  of  any  quantity  is  plus.     Hence,  since 
the  operations  leave  the  inequality  in  the  same  sense  as  the  first, 

3.  Show  that  a2  -f  b2  +  c2  >  afr  +  ac  4-  6c. 
Solution.     By  Ex.  1,  a2  +  b2>2ab, 

a2  +  c2  >  2  ac, 
62  +  c2  >  2  6c. 
Adding  and  dividing  by  2,  a2  -f  62  4  c2  >  «6  +  ac  4  6c. 

4.  Which  is  greater,  — or ,  if  x  <  a  ? 

a-  +  r        a-\-x 

5.  Show  that  any  fraction  plus  its  reciprocal  is  greater  than  2. 

6.  If  a,  b,  c  are  such  that  the  sum  of  any  two  is  greater  than 
the  third,  show  that  a2  -f  b2  -f-  c2  <  2  (a&  -4-  ac  4-  be). 

7.  If   a2  +  b2  +  c2  =  1,    and   m2  +  n2  +  r2  =  1,    show   whether 
am  +  £m  -f  cr  is  greater  or  less  than  1. 

8.  Show  that 

(a  +  b  —  c)2+(a  +  c  —  b)2  +  (6  +  c  —  a)2  >  a/>  4-  be  +  ac. 

9.  Which  is  greater,  a3  4-  &3  or  a2£>  +  a&2  ? 

10.  Prove  that  (a&  4-  xy)  (ax  +  %)  >  4  abxy. 

11.  Which  is  greater,  2  Xs  or  sc  4-  1,  if  x  >  1  ? 

12.  Find  the  limits  of  x  determined  by  the  conditions 

x  —Z  .  x  —  4  .    A         -,  a;  —  10    -  x  -\-  5      1 

>  9,  and  <  — — 

95'  69         6 


INEQUALITIES  173 

13.  The  double  of  a  number  diminished  by  5  is  greater  than 
25,  and  the  triple  of  the  number  diminished  by  7  is  less  than 
the  double  increased  by  13.     Find  the  limits  of  the  number. 

14.  A  man  wishes  to  make  a  purchase  for  $  14,  but  has  not 
enough  money.  If  he  borrows,  one  third  as  much  money  as  he 
now  has,  he  will  be  able  to  make  the  purchase  and  have  more 
money  left  than  he  now  lacks.     How  much  money  has  he  ? 

15.  The  daily  pay  roll  of  a  contractor,  who  pays  masons  $  4.40 
a  day  and  carpenters  $  3.60  a  day,  is  between  $  104  and  $  112, 
and  there  are  3  more  masons  than  carpenters.  Find  the  number 
of  each. 

16.  A  is  24  years  old  and  B  is  15.  What  is  the  shortest  time 
after  which  A's  age  will  be  less  than  If  times  B's  age  ? 

17.  If  only  5  pupils  be  seated  on  each  bench  in  a  recitation 
room  containing  fewer  than  6  benches,  4  pupils  will  be  without 
seats ;  but  if  6  pupils  be  seated  on  each  bench,  some  seats  will  be 
unoccupied.     Find  the  number  of  benches. 

18.  The  sum  of  two  whole  numbers  is  25.  If  the  greater  be 
divided  by  the  less,  the  quotient  will  be  less  than  3J ;  and  if  the 
less  be  divided  by  the  greater,  the  quotient  will  be  greater  than  £. 
What  are  the  numbers  ? 


CHAPTER   XIII 

RATIO,    PROPORTION,    AND    VARIATION 

SECTION   I  — RATIO 

276.  Ratio  is  the  relative  magnitude  of  two  quantities  of  the 
same  kind,  and  is  measured  by  the  quotient  of  the  first  by  the 
second. 

The  ratio  of  two  quantities  is  expressed  either  by  writing  a 
colon  between  them  or  by  writing  them  in  the  fractional  form. 

Thus  the  ratio  of  a  to  6  is  written  either  a  :  b  or  -• 

b 

277.  The  Antecedent  or  First  Term  of  a  ratio  is  the  first  of  the 
two  quantities  compared,  and  the  Consequent  or  Second  Term  is 
the  second. 

278.  A  ratio  is  a  Ratio  of  Greater  Inequality,  Less  Inequality, 
or  Equality,  according  as  it  is  greater  than,  less  than,  or  equal  to, 
unity. 

279.  The  Duplicate,  Sub-duplicate,  Triplicate,  and  Sub-triplicate 
Ratios  of  two  quantities  are  the  ratios  of  the  squares,  square 
roots,  cubes,  and  cube  roots  of  those  quantities  respectively. 

280.  A  Compound  Ratio  is  the  ratio  of  the  products  of  the  cor- 
responding terms  of  two  or  more  simple  ratios. 

281.  Theorem.  1st.  A  ratio  is  not  changed  by  multiplying  or 
dividing  both  its  terms  by  the  same  quantity. 

2d.  A  ratio  is  multiplied  by  multiplying  its  antecedent  or  dividing 
its  consequent. 

3d.  A  ratio  is  divided  by  dividing  its  antecedent  or  multiplying 
its  consequent. 

174 


RATIO  175 

Since  a  ratio  is  simply  a  fraction,  or  an  indicated  division, 
these  follow  from  Arts.  127  and  137. 

282.  Theorem.  A  ratio  of  greater  inequality  is  diminished,  and 
a  ratio  of  less  inequality  is  increased,  by  adding  the  same  positive 
quantity  to  both  its  terms;  i.e.,  any  ratio  is  made  more  nearly  equal 
to  unity  by  adding  the  same  positive  quantity  to  both  its  terms. 

Dem.     Let  m :  n,  or  — ,  be  the  given  ratio,  and  a  the  quantity 
n 

to  be  added  to  both  its  terms.     Then  we  have 

m  >  m  +  a  ^ . 

n  <  n+a'  ' 

Whichever  member  is  greater  will  still  be  greater  after  both 
are  multiplied  by  n  (n  +  a),  giving 

mn  +  am  ^  mn  +  an.  (2) 

Since  mn  is  common  to  both  members,  the  first  member  of  (2) 
will  be  greater  or  less  than  the  second,  according  as  am  is  greater 
or  less  than  an,  or  m  greater  or  less  than  n.  Hence  the  first 
member  of  (1)  is  greater  or  less  than  the  second,  according  as  m 
is  greater  or  less  than  n. 


EXAMPLES  LXXVI 
Find  the  values  of  the  following  ratios ; 
1.  x2  —  7  x  +  10:  x  —  2.  2.   #+5:o2  +  3a;-10. 

3.   x*-y*:xi-x2y4-xy2-y\  4.   12  (a  -  b)2 :  8  (a2  -  b2). 

5.  3x3-2x2-19x-6:3xi  +  4:X2-5x-2. 

6.  The  duplicate  ratio  of  13  :  39. 

7.  The  sub-duplicate  ratio  of  121^-726  a; +1089  :  a^-Gz+a 

8.  The  triplicate  ratio  of  } :  | • 

9.  The  sub-triplicate  ratio  of  729  :  1728. 

10.  Is  a2  —  x2 :  a2  +  x2  greater  or  less  than  a  —  x.a  +  x  ? 

11.  Is  x3  +  i/3  :  x2  +  y2  greater  or  less  than  x2  +  y2 :  x  +  y  ? 


176  HIGHER  ALGEBRA 

12.  Is  Xs  —  y* :  (x  4-  yf  greater  or  less  than  x2  -f-  xy  -f-  y2 :  x  —  #, 
a;  being  greater  than  y  ? 

13.  If  6  be  added  to  each  of  two  numbers  that  are  in  the  ratio 
of  5  :  7,  the  sums  will  be  in  the  ratio  of  7:9.     Find  the  numbers. 

14.  If  8  be  subtracted  from  each  of  two  numbers  that  are  in 
the  ratio  of  2:5,  the  remainders  will  be  in  the  ratio  of  2:9. 
Find  the  numbers. 

15.  What  must  be  subtracted  from  each  term  of  13  :  25  to  make 
it  1:3? 

16.  A  certain  ratio  becomes  J  when  4  is  added  to  both  its  terms, 
and  I-  when  2  is  subtracted  from  both  its  terms.     Find  the  ratio. 

17.  What  quantity  subtracted  from  each  term  of  the  duplicate 
ratio  of  m :  n  will  give  the  triplicate  ratio  of  m  :  n  ? 

18.  If  5  gold  coins  and  30  silver  ones  are  worth  as  much  as  10 
gold  coins  and  10  silver  ones,  what  is  the  ratio  of  their  values  ? 

19.  If  the  wages  of  5  boys  and  6  girls  is  J  of  the  wages  of  6  boys 
and  9  girls  for  the  same  time,  what  is  the  ratio  of  their  wages  ? 

20.  The  sides  of  a  triangle  are  in  the  ratio  of  3,  4,  and  5,  and 
the  perimeter  is  480  yards.     Find  the  sides. 

21.  The  ratio  of  a  father's  age  to  his  son's  is  7  :  2,  and  the  father 
is  30  years  older  than  the  son.     Find  the  age  of  each. 

22.  A  fox  makes  4  leaps  while  a  hound  makes  3 ;  but  2  of  the 
hound's  leaps  equal  3  of  the  fox's.    Find  relative  rates  of  running. 

23.  If  in  working  out  road  tax  6  men  and  4  teams  are  counted 
as  much  as  10  men  and  1  team,  what  is  the  ratio  of  wages  for 
men  and  teams  ? 

24.  A  bequest  of  $  900  wa,s  divided  among  three  sons,  and  after 
their  shares  had  increased  by  $  10,  $  15,  and  $  20  respectively, 
the  sums  were  in  the  ratio  4:5:6.     Find  the  shares. 

25.  The  weights  of  two  loads  are  in  the  ratio  of  4  to  5.  Parts 
of  the  loads  in  the  ratio  of  6  to  7  being  removed,  the  remaining 
weights  are  in  the  ratio  of  2  to  3,  and  the  sum  of  the  weights  is 
then  10  tons.     What  were  the  weights  at  first  ? 


Piwpoinio.x  177 

26.  In  a  college  boat  race,  crew  A  pull  15  strokes  to  14  strokes 
of  crew  B ;  but  28  strokes  of  crew  B  are  as  effective  as  33  of  crew 
A.     Which  is  the  faster  crew,  and  in  what  ratio  ? 

27.  Find  the  gear  of  a  bicycle  whose  wheels  are  d  inches  in 
diameter,  whose  front  sprocket  has  m  teeth  and  whose  rear  sprocket 
li;is  n  teeth,  it  being  understood  that  the  "gear''  of  a  bicycle  is 
the  diameter  of  a  wheel  one  revolution  of  which  would  advance  it 
as  far  as  one  revolution  of  the  pedal  advances  the  bicycle. 

Solution.     Let  x  =  the  gear. 

One  revolution  of  a  wheel  whose  diameter  is  x  would  advance  the  wheel  by 
the  amount  of  its  circumference,  ttx. 

The  circumferences  of  the  sprocket  wheels  are  in  the  ratio  of  their  number  of 

teeth,  m  :  n.     Hence  —  =  the  number  of  revolutions  of  the  bicycle  wheels  to 

M 
one  revolution  of  the  pedal,  and  -  x  wd  =  the  distance  the  bicycle  advances 

for  one  revolution  of  the  pedal.     Therefore, 

m    7 
irx  =  —  ?rf7, 
n 

md 
whence  x  = 

N 

28.  Find,  by  substituting  in  the  formula  of  the  last  example, 
the  gear  in  each  of  the  following  cases : 

(a)  d  =  2S  in.,  m  =  18,  n  =  7. 

(b)  d  =  2S  in.,  m  =  21,  n  =  7. 

(c)  d  =  30  in.,  m  =  20,  n  =  8. 

(d)  d  =  28  in.,  m  =  22,  n  =  8. 

SECTION  II  — PROPORTION 

283.    A  Proportion  is  an  equality  of  ratios. 

The  equality  is  indicated  by  the  sign  of  equality  or  by  the 
double  colon.  If  the  ratio  a  :  b  equals  the  ratio  c :  d,  the  propor- 
tion may  be  written  in  any  one  of  the  three  ways, 

a:b: :  c  :  d,  a  :  b  —  c  :  d,  or  -  =  -• 
'        b      d 

In  any  form  the  proportion  is  read,  "  a  is  to  b  as  c  is  to  pL" 
downey's  alo.  — 12 


178  ITTGTIER   ALGEBRA 

284.  Four  quantities  are  Directly  Proportional  when  the  ratio  of 
two  of  them  is  equal  to  the  ratio  of  the  other  two  taken  in  the 
same  order. 

Thus,  the  times  being  the  same,  any  two  distances  are  directly  propor- 
tional to  the  corresponding  rates. 

285.  Four  quantities  are  Inversely  or  Reciprocally  Proportional 
when  the  ratio  of  two  of  them  is  equal  to  the  ratio  of  the  other 
two  taken  in  the  inverse  order. 

Thus,  the  distances  being  the  same,  any  two  times  are  inversely  or  recip- 
rocally proportional  to  the  rates.  If  T  is  the  time  at  the  rate  li,  and  t  the 
time  at  the  rate  r,  then 

T:t::r:B,   or  T :  t  :  :  —   :  -• 
li      r 

The  same  relation  is  expressed  by  saying,  "The  times  are  in  the  inverse, 
or  reciprocal,  ratio  of  the  rates,"  or  "The  times  vary  inversely,  or  recipro- 
cally, as  the  rates." 

286.  The  Extremes  of  a  proportion  are  its  first  and  last  terms. 
The  Means  of  a  proportion  are  its  second  and  third  terms. 

287.  When  the  means  of  a  proportion  are  the  same  quantity, 
this  quantity  is  called  a  Mean  Proportional  between  the  other  two 
quantities,  and  the  last  term  is  called  a  Third.  Proportional  to  the 
other  two  quantities. 

Thus,  in  a  :  b  :  :  b  :  c,  b  is  a  mean  proportional  between  a  and  c,  and  c  is  a 
third  proportional  to  a  and  b. 

288.  A  Fourth  Proportional  to  three  quantities  is  the  fourth  term 
of  a  proportion  whose  other  three  terms  are  the  three  quantities 
taken  in  their  order. 

289.  A  proportion  is  taken  by  Inversion  when  the  terms  of 
each  ratio  are  written  in  inverse  order. 

290.  A  proportion  is  taken  by  Alternation  when  the  means  are 
interchanged,  or  when  the  extremes  are  interchanged. 

291.  A  proportion  is  taken  by  Composition  when  the  sum  of  the 
terms  of  each  ratio  is  compared  with  either  term  of  that  ratio,  the 


PROPORTION  179 

same  order  being  observed ;  or  when  the  sum  of  the  antecedents 
and  the  sum  of  the  consequents  are  compared  with  either  ante- 
cedent and  its  consequent. 

292.  A  proportion  is  taken  by  Division  if  difference  be  substi- 
tuted for  sum  in  the  last  definition. 

293.  A  Continued  Proportion  is  a  succession  of  equal  ratios  in 
which  each  consequent  is  the  antecedent  of  the  next  ratio. 

Thus,  a  :b  :  :b  :  c:  :  c  :d:  :  d  :  e  is  a  continued  proportion. 

294.  Theorem.  In  any  proportion  the  product  of  the  extremes 
equals  the  product  of  the  means. 

Dem.     Let  the  proportion  be 

a  :  b  :  :  c  :  d. 

This  is  the  same  as  ^  =  -  (Art.  283). 

o      a 

Clearing  of  fractions,  ad  =  be. 

295.  Cor.  i.  A  mean  proportional  between  two  quantities  is 
equal  to  the  square  root  of  their  product. 

For  if  a:b:  :b:c, 

b2  =  ac, 
whence  b  =  sfac. 

296.  Cor.  2.  Either  extreme  of  a  proportion  equals  the  product 
of  the  means  divided  by  the  other  extreme;  and  either  mean  equals 
the  product  of  the  extremes  divided  by  the  other  mean. 

297.  Theorem.  If  the  product  of  two  quantities  equals  the  prod- 
uct of  two  others,  the  quantities  of  one  product  may  be  made  the 
extremes  and  the  quantities  of  the  other  product  the  means  of  a 
proportion. 

Dem.     Let  ad  =  be. 

Dividing  by  M,  £=|; 

o     a 

that  is,  a  :  b  : :  c  :  d. 


180  HIGHER  ALGEBRA 

Writing  the  first  equation  in  the  form 

bc  =  ad, 

and  dividing  by  ac,  -  =  -; 

that  is,  b:a::  d:c. 

By  dividing  by  other  combinations  of  the  letters  other  forms 
can  be  obtained ;  but  in  each  the  quantities  of  one  product  will  be 
the  extremes,  and  the  quantities  of  the  other  product  the  means. 

298.  Theorem.  Proportionals  result  from  taking  a  proportion 
(a)  by  inversion,  {b)  by  alternation,  (c)  by  composition,  (d)  by 
division,  (e)  by  composition  and  division. 

Dem.     Let  the  given  proportion  be  * 

a:b  ::  c:  d.  (1) 

Then  proportionals  result  from  taking  this 


(2) 
(3) 


(a)  By  inversion, 

b:  a:  :  d:  c. 

From  (1), 

a  _c 
b~d 

Dividing  unity  by  each  member, 

b  __d. 

—     5 
a     c 

that  is, 

b:  a:  :  d:  c. 

(b)  By  alternation, 

(  a:  c:  :b:  d, 
[d:  b:  :  c  :  a. 

From  (1), 

ad  =  be. 

Dividing  by  cd, 

a      b 
c~d' 

that  is, 

a:c:  :b:d. 

Dividing  (4)  by  ab, 

d _c  m 
b      a' 

that  is, 

d:  b: :  c:  a. 

(4) 


PROPORTION 


181 


or 


'  a-\-  b  :  b  : :  c  -\-d 

4 

a  +  b'.awc  +  d: 

c, 

(c)  By  composition, 

a  +  c:  a:  :b  +  d 

&, 

a  +  c  :  c  : :  b  +  d: 

d, 

and  other  forms. 

Adding  unity  to  both  members  of  (2), 

-  +  1=-+1, 

a  +  b  _c  +  d , 
b            d 

(5) 


that  is,  a  +  6  :  b  : :  c  +  d  :  d. 

Let  the  student  demonstrate  the  other  three  forms  given. 
Other  forms  are  obtained  by  taking  these  four  forms  by  inversion 
and  by  alternation. 


or 


r  a  —  b  :  b  :  :  c  —  d  :  d, 

a  —  b  :a::  c  —  d:  c, 

(d)  By  division, 

a  —  c  :  a :  :  b  —  d  :  b, 

a  —  c  :  c  :  :  b  —  d  :  d, 

and  other  forms. 

Subtracting  unity  from  both  members  of  (2), 

«-l=c--l, 
b            d 

a—b     c—d 

b            d     ' 

at  is,                             c 

i  —  b:b  :  :c  —  did. 

(6) 


Let  the  student  demonstrate  the  other  three  forms  given. 
Other  forms  are  obtained  by  taking  these  four  forms  by  inversion 
and  by  alternation. 

(  a  +  b  :  a  —  b  :  :  c  +  d:  c  —  d, 

]  a  -f-  c:  a  —  c::b  +  d:l 
and  other  forms. 
a  +  b  _c  +  d.m 
a  —  b      c  —  d1 
that  is,  a  -\-  b  :  a  —  b  :  \  c  -\-  d  :  c  —  d. 


(e)  By  composition 
and  division, 

Dividing  (5)  by  (0), 


d, 


182  HIGHER  ALGEBRA 

Let  the  student  demonstrate  the  othor  form  given.  Other 
forms  are  obtained  by  taking  these  two  forms  by  inversion  and 
by  alternation. 

299.  Theorem.  If  four  quantities  are  in  proportion,  proportionals 
result  from  taking  equimultiples,  (a)  of  the  terms  of  a  couplet,  (b)  of 
the  antecedents,  (c)  of  the  consequents,  (d)  of  all  the  terms. 

Dem.     Let  the  given  proportion  be 

a :  b : :  c :  d,  (1) 

Then  proportionals  result  from  taking  equimultiples, 
(a)   Of  the  terms  of  a  couplet. 

From(l),  l  =  c- 

o      d 

Multiplying  numerator  and  denominator  of  the  first  member 

by  m, 

am  _c  m 
bm      d' 

that  is,  am:bm::c:  d. 

Let  the  student  demonstrate  the  other  cases. 

300.  Theorem.  The  same  powers  or  roots  of  proportionals  are 
proportional. 

Let  the  student  demonstrate. 

301.  Theorem.  The  products  or  the  quotients  of  the  corresponding 
terms  of  two  (or  more)  proportions  are  proportional. 

Dem.     Let  the  given  proportions  be 

a  :  b  : :  c  :  d, 


a) 

(2) 


and 

m:n::p:  q. 

These  are  the  same  as 

a  _c 
b~d 

and 

m      p 
n      q 

PROPORTION  183 


Multiplying  (1)  by  (2),       j^  =  %\ 

that  is,  am  :bn  : :  ep  :  dg« 

From  the  given  proportions, 


Dividing  (3)  by  (4), 


or 


ad  -- 

=  be, 

mq  -- 

=  np. 

ad 
mq 

_bc 

~ np 

a      d 

—  X  -  = 
m      q 

b      c 

=  -  x  - 

n     p 

a    b 

c    d 

m'  n ' 

"  p'q 

(3) 

(4) 


Hence,  by  Art.  297, 


302.  Theorem.  In  a  series  of  equal  ratios  the  sum  of  all  the  ante- 
cedents is  to  the  sum  of  all  the  consequents  as  any  antecedent  is  to  its 
consequent. 

Dem.     If  a:b  ::  c:  d::  e  :  f::  g :  h,  etc.,  (1) 

then     a  +  c  +  e+g  +  etc. :  b  +  d +f-\-h  +  etc.  : :  a  :  b  or  c :  d,  etc. 
We  have  the  identity,  ab  =  ba. 

From  (1)  we  have  ad  =  be, 

af=be, 

ah  =  bg, 

etc.,    etc. 

Adding,  a{b  +  d  +/-f  h  +  etc.)=  b{a  +  c  +  e  +  g  +  etc.). 
Hence,  by  Art.  297, 

a  +  c  +  e  +  g  +  etc.  :b  +  d  +f+ h  +  etc.  ::a:b. 

303.  When  a  proportion  is  given  and  we  wish  to  determine 
whether  some  other  proposed  relation  involving  the  same  quan- 
tities is  true,  we  may  proceed  in  any  one  of  several  ways.  For 
example,  let  the  given  proportion  be 

a  :  b  :  :  c  :  d, 

and  let  the  problem  be  to  determine  whether 

a  -f-  b  :  b  : :  c  -f  d  :  d. 


184  HIGHER  ALGEBRA 

1st.  We  may  proceed  as  in  the  demonstration  of  this  case, 
Art.  198,  c. 

2d.    The  proposed  proportion  is  true  if  it  can  be  shown  that  the 

product  of  the  extremes  equals  the  product  of  the  means.     This 

would  give 

ad  -f  bd  =  be  -J-  bd, 

or  ad  =  be, 

which  is  seen  from  the  given  proportion  to  be  true. 

3d.  In  the  given  proportion  we  may  represent  each  of  the 
equal  ratios,  a :  b  and  c :  d,  by  r,  giving 

a  -,    c 

-  =  r    and   -  =  r, 

b  d 

or  a  =  br  and  c  =  dr. 

Substituting  these  values  in  the  proposed  form,  we  have 

br  +  b :  b : :  dr  +  d:d, 

in  which  the  ratios  are  seen  to  be  equal,  each  being  r  +  1  : 1. 

Any  of  the  forms  of  the  preceding  theorems  may  be  tested  in 
this  way. 

EXAMPLES  LXXVII 

1.  Find  the  ratio  of  x  to  y  in  7  x  —  5  y  :  4:  x  —  3  y: :  5  :  2. 

2.  "From  a:b::  c  :  d  deduce 

5  a  +  S  b :  5  a  —  3  & : :  &c  +  3  d :  5  c  —  3  d 

3.  From  x  :  y  : :  4  :  7,  find  the  ratio  of  #  —  4  to  y  —  7. 

4.  Find  the  mean  proportional  between  x2 and  if  — -- 

y  *^ 

5.  Find  x  from  x2+5  x+6 :  a^  +  lO  tf+21 : :  x2-± :  4^+8  a;- 32. 

6.  The  third  proportional  to  two  numbers  is  48,  and  the  mean 
proportional  between  them  is  6.     Find  the  numbers. 

7.  If  a,  b,  c,  d  are  in  continued  proportion,  prove  that  b  +  c  is 
a  mean  proportional  between  a  -f  6  and  c  +  c/. 

8.  From  a:b  ::  c:d  deduce  a—  c:  b  —  d::  Va2  -f-  c* :  V&2  +  d2. 

9.  From  |a-x;ia  +  a;::6-?/:6  +  ?/  deduce  2  x\  y:\a\b. 
Sue.     Take  by  division  and  composition,  divide  second  couplet  by  2,  and 

take  by  alternation. 


PROPORTION  185 

10.  If  a : b : : e:d : : e:f,  prove  that 

a  +  3  c  +  2  e  :  a  -  e  : :  6  +  3  d  +  2/:  6  -/. 

11.  What  operations  on  x  :  y  : :  a  :  b  will  produce 

sc2 :  a2 : :  a2  -f  y2 :  a2  +  62  ? 

12.  If  a  :  6  : :  »  :  q,  prove  that  a2  +  b2 : : :  p2  +  q2 :  — 

a  +  b  p  +  q 

13.  What  quantity  added  to  each  of  the  quantities  a,  b,  c,  dy 
will  make  them  proportionals  ? 

14.  If  four  quantities  are  proportionals,  show  that  there  is  no 
quantity  which,  being  added  to  each,  will  leave  the  sums  propor- 
tionals. 

15.  If  the  fourth  proportional  to  a,  b,  c  is  the  same  as  that  to 
a,  b'j  c',  show  that  b  :  b' : :  c' :  c. 

16.  Iia:b::c:d,  prove  that  a2  +  b2 :  c2  +  d2 :  :  (a  +  &)2 :  (c  +  tf)2- 

17.  If  a :  b : :  c :  d,  prove  that  ab  -f  cd  is  a  mean  proportional 
between  a2  +  c2  and  b2  +  d2- 

18.  What  quantity  subtracted  from  each  of  the  quantities 
a,  6,  c  will  leave  the  remainders  in  continued  proportion  ? 

19.  If  x  be  to  y  in  the  duplicate  ratio  of  a  to  6,  and  a  be  to  b 
in  the  sub-duplicate  ratio  of  a  +  x  to  a  —  y,  prove  that 

2  x  :  a  :  :  x  —  y  :  y. 

20.  A  farmer's  crop  of  wheat  was  to  his  crop  of  oats  as  2:3. 
His  neighbor  raised  50  bushels  more  of  each,  and  his  crop  of 
wheat  was  to  his  crop  of  oats  as  5:7.  How  many  bushels  of 
each  did  the  first  farmer  raise? 

Sug.  If  in  solving  such  problems  proportions  are  used,  place  die  product 
of  the  extremes  in  each  equal  to  the  product  of  the  means,  and  solve  the 
equations  in  the  usual  way.  In  this  example  but  one  unknown  quantity  is 
necessary,  as  we  may  represent  by  2  x  and  3  x  the  number  of  bushels  of  wheat 
and  oats  respectively. 

21.  Divide  14  into  two  such  parts  that  the  greater  divided  by 
the  less  shall  be  to  the  less  divided  by  the  greater  as  16  to  9. 


186  HIGH  Eli  ALGEBRA 

22.  Two  vessels  contain  respectively  15  and  27$  gallons.  How 
many  gallons  must  be  transferred  from  one  to  the  other  that  the 
amounts  may  be  in  the  ratio  2:3? 

23.  It  is  required  to  find  a  number  such  that  the  sum  of  its 
digits  is  to  the  number  itself  as  4  to  13,  and  the  difference  of  its 
digits  is  to  the  number  expressed  when  the  digits  are  inter- 
changed as  2  to  31. 

24.  Two  numbers  having  the  same  two  digits  are  to  each  other 
as  5  :  6.     What  are  the  numbers  ? 

25.  Find  two  numbers  such  that  their  sum,  difference,  and 
product  may  be  as  the  numbers  «s,  d,  and  p,  respectively. 

26.  Find  two  numbers  whose  difference  is  to  the  difference  of 
their  squares  as  m :  n,  and  whose  sum  is  to  the  difference  of  their 
squares  as  a :  b. 

27.  A  bin  contains  chop-feed  composed  of  corn  and  oats.  A 
second  bin  contains  6  bushels  more  of  each,  and  has  7  bushels  of 
corn  to  every  6  bushels  of  oats.  A  third  bin  contains  6  bushels 
less  of  each  than  the  first,  and  has  6  bushels  of  corn  to  every  5 
bushels  of  oats.  How  many  bushels  of  each  does  the  first  bin 
contain  ? 

28.  The  force  of  the  earth's  attraction  is  inversely  as  the  square 
of  the  distance  from  the  center.  At  the  surface  this  force  is  ex- 
pressed by  the  number  32.16.  By  what  is  it  expressed  at  the 
moon,  whose  distance  from  the  center  of  the  earth  is  60  radii  of 
the  earth  ? 

29.  The  velocities  of  bodies  revolving  around  another  body  are 
inversely  proportional  to  the  squares  of  the  distances.  If  the 
velocity  is  v  when  the  distance  is  r,  what  is  it  when  the  dis- 
tance is  r'  ? 

30.  Two  men  have  equal  capital  in  business,  and  one  is  losing 
as  fast  as  the  other  is  gaining.  A  third  man's  capital  is  less  byp 
dollars,  but  is  rapidly  increasing.  Show  that  if  m  and  n  are  the 
respective  gains  of  the  third  man  at  the  times  when  he  has  the 
same  amounts  as  the  others,  then  2  mn  =_p(m  -f-  ri). 


VARIATION  187 

31.  Before  noon,  a  clock  which  is  too  fast  and  points  to  after- 
noon time,  is  turned  back  5  hours  and  40  minutes  to  the  true 
time;  and  it  is  observed  that  the  time  before  shown  is  to  the 
true  time  as  29  to  105.     Find  the  true  time. 

32.  Areas  of  circles  are  as  the  squares  of  their  diameters. 
Two  circular  metallic  plates,  each  an  inch  thick,  whose  diameters 
are  6  and  8  inches  respectively,  are  melted  and  cast  into  a  single 
circular  plate,  1  inch  thick.     Find  its  diameter. 

33.  The  volumes  of  spheres  are  as  the  cubes  of  their  radii. 
Find  the  radius  of  a  sphere  whose  volume  is  equal  to  the  sum  of 
the  volumes  of  three  spheres  whose  radii  are  3,  4,  and  5,  re- 
spectively. 

34.  The  rates  of  an  express  train  and  an  accommodation  train 
differ  by  15  miles  an  hour,  and  their  times  of  making  a  distance 
of  180  miles  are  as  9  :  14.  The  express  train  loses  by  stoppages 
only  half  as  much  time  as  the  accommodation  train,  and  the 
latter  thus  loses  as  much  time  as  it  would  require  in  running  30 
miles.     Find  the  rates. 

35.  Two  passengers  have  together  560  pounds  of  baggage,  and 
are  charged  for  the  excess  above  the  weight  allowed  62  cents  and 
$  1.18  respectively.  A  third  passenger  has  as  much  baggage  as 
the  other  two,  and  is  charged  $  2.30  for  excess.  How  much 
baggage  is  each  passenger  allowed  without  charge  ? 

SECTION  III  — VARIATION 

304.  When  the  ratio  of  two  quantities  is  constant,  either  is 
said  to  vary  directly  as  the  other,  or  simply  to  vary  as  the  other. 

Thus,  if 

x  =  my, 

in  which  m  is  constant,  any  change  in  y  causes  x  to  change  in  the  same  ratio : 
if  y  be  doubled,  x  will  be  doubled  ;  if  y  be  tripled,  x  will  be  tripled  ;  if  y  be 
halved,  x  will  be  halved  ;  and  so  on. 

This  relation  between  x  and  y  is  written 

XQC?/, 

and  is  read  "  x  varies  as  y." 


188  HIGHER  ALGEBRA 

Distance  traveled  at  a  given  rate  varies  as  the  time,  the  constant  ratio 

being   the   given   rate.     If  d,  r,  and  t  represent  distance,  rate,  and  time 

respectively, 

d  =  rt, 

and  dcct. 

Amount  earned  in  a  given  time  varies  as  the  daily  wages,  the  constant  ratio 

being  the  given  time.     If  a,  t,  and  ic  represent  amount,  time,  and  daily  wages 

respectively, 

a  =  tw, 

and  a  x  w. 

The  circumference  of  a  circle  varies  as  its  radius,  the  constant  ratio  being 
2  7r.     If  c  and  r  represent  the  circumference  and  radius  respectively, 

c  =  2  7rr, 
and  cccr. 

305.  When  the  ratio  of  a  quantity  to  the  reciprocal  of  another 
is  constant,  either  is  said  to  vary  inversely  as  the  other. 

'     ThUS'if  *=* 

in  which  m  is  constant,  any  change  in  y  causes  x  to  change  in  the  inverse  ratio 
(Art.  285)  :  if  y  be  doubled,  x  will  be  halved  ;  if  y  be  tripled,  x  will  be  made 
one  third  as  great ;  if  y  be  halved,  x  will  be  doubled  ;  and  so  on. 
This  relation  between  x  and  y  is  written 

20C1, 

y 

and  is  read,  "  x  varies  inversely  as  ?/." 

The  time  required  to  travel  a  given  distance  varies  inversely  as  the  rate, 
the  constant  ratio  being  the  given  distance.  If  t,  d,  and  r  represent  time, 
distance,  and  rate  respectively, 


and  t  x  - . 

r 

The  time  req  lired  to  earn  a  given  amount  varies  inversely  as  the  daily 
wages,  the  constant  ratio  being  the  given  amount.  If  t,  d,  and  w  represent 
time,  amount,  and  daily  wages  respectively, 


w 

and  U-, 

w 


VARIATION  189 

306.  When  the  ratio  of  one  quantity  to  the  product  of  two  or 

more  other  quantities  is  constant,  the  first  is  said  to  vary  jointly 

as  the  others. 

Thus,  if 

x  =  myz, 

in  which  m  is  constant,  any  change  in  y  and  z  causes  x  to  change  in  a  ratio 
represented  by  the  product  of  these  changes  ;  if  y  be  doubled  and  z  tripled, 
x  will  be  made  six  times  as  great ;  and  so  on. 
This  relation  between  x  and  yz  is  written 

xctyz, 

and  is  read,  "x  varies  jointly  as  y  and  z." 

Distance  traveled  varies  jointly  as  the  time  and  rate,  the  constant  ratio 
being  1. 

The  area  of  a  triangle  varies  jointly  as  its  base  and  altitude,  the  constant 
ratio  being  \.     If  s,  b,  and  h  represent  area,  base,  and  altitude  respectively, 

s=  \bh, 
and  s  x  bh. 

307.  When  the  ratio  of  one  quantity  to  the  quotient  of  a 
second  divided  by  a  third  is  constant,  the  first  is  said  to  vary 
directly  as  the  second  and  inversely  as  the  third. 

Thus,  if 


V 
x  —  m—, 

z 


in  which  m  is  constant,  then 


1 
which  is  read,  "x  varies  directly  as  y  and  inversely  ass."' 

If  time,  distance,  and  rate  are  all  variable,  the  time  varies  directly  as  the 
distance  and  inversely  as  the  rate. 

In  doing  work,  the  time  varies  directly  as  the  amount  of  work  and 
inversely  as  the  number  of  workmen  employed. 

308.    Theorem.     If  one  set  of  corresponding  values  of  the  vari- 
ables of  a  variation  be  given,  the  constant  ratio  becomes  known. 
Dem.     Let  the  variation  be 

xcKy.  (1) 

By  definition  (Art.  304)  the  ratio  of  x  to  y  is  constant.     Let 
this  constant  ratio  be  m.     Then 

x  ^  my.  (2) 


190  HIGHER  ALGEBRA 

Or  we  may  say,  since  x  varies  as  y,  x  is  always  a  certain  num- 
ber of  times  y,  as  m  times  y. 

Let  a  and  b  be  corresponding  values  of  x  and  y.     Substituting 

these  in  (2),  we  have 

a  =  mb, 

whence  m  =  — 

b 

The  same  reasoning  applies  to  the  other  forms  of  variation. 

309.  Cor.  To  pass  from  a  variation  to  an  equation,  a  constant 
factor,  the  ratio,  is  introduced;  and,  conversely,  to  pass  from  an 
equation  to  a  variation,  constant  factors  are  omitted. 

310.  Caution.  When  two  or  more  variations  occur  in  the 
same  problem,  the  same  constant  must  not  be  used  twice  in  pass- 
ing from  variations  to  equations ;  for,  while  in  each  variation  the 
ratio  is  constant,  it  will  not  do  to  assume  that  it  is  the  same. 

311.  Theorem.  A  variation  may  always  be  expressed  as  a  pro- 
portion, and  is,  in  fact,  simply  a  contracted  proportion. 

Dem.     By  definition  the  expression 

xocy 
signifies  that  whatever  value  x  may  have,  y  has  a  corresponding 
value,  such  that  the  ratio  of  x  to  y  is  constant.     Letting  x'  and  x" 
be  two  values  of  x,  and  y'  and  y"  the  corresponding  values  of  y, 
the  ratio  x' :  y'  is  the  same  as  the  ratio  x"  :  y";  that  is, 

x' :  y' :  :  x"  :  y", 
or,  if  we  choose  (Art.  298,  b), 

x'  :x"::y':  y". 
The  same  reasoning  applies  to  the  other  forms  of  variation. 

EXAMPLES   LXXVHI 

1.   If  x  oc  y  and  y  cc  -,  show  how  x  varies  with  reference  to  z. 

z 


Solution.     Since 

xccy, 

we  have  (Art.  309) 

x  =  my; 

and  since 

ycc-> 
z 

we  have 

n 

y  =  r 

(1) 

(2) 


VA  It  I  ATI  ON  191 

Eliminating  y  from  (1)  and  (2), 

mn 

whence  (Art.  309)  %  «-, 

2.  If  x  oc  0  and  ?/  oc  z,  show  how  #  varies  with  reference  to  ?/. 

3.  If  x  oc  2/2  and  y  oc  z2,  show  how  z  varies  with  reference  to  z. 

4.  If  x2  x  z  and  ?/2  oc  z,  prove  that  a#  ac  z. 

5.  If  a*  oc  y  and  g  =  14  when  y  =  §,  what  is  the  value  of  g  in 
terms  of  y  ? 

Solution.     Since  xxj/, 

we  have  (Art.  300)  x  =  my. 

Substituting  the  given  corresponding  values, 

14  =  %  m, 
whence  m  =  35. 

Since  m  is  constant,  when  we  have  found  its  value  for  one  pair  of  corre- 
sponding values  of  x  and  ?/,  we  have  it  for  all  corresponding  values  of  x  and 

y.     Hence 

x  =  35?/. 

6.  If  a;  oc  y,  and  x  =  20  when  y  =  2\,  what  is  the  value  of  x  in 
terms  of  y  ? 

7.  If  x  oc  -,  and  x  =  13  when  y  =  4,  what  is  the  value  of  x  in 

y 
terms  of  y  ? 

8.  If  a;  oc  — ,  and  x  =  8  when  y  =  h  what  is  the  value  of  x  in 

y2 
terms  of  ?/  ? 

9.  If  #  oc  y,  and  x  =  V2  when  2/  =  3,  what  is  the  value  of  y 
when  a;  =  32  ? 

1st  Solution.    Since  xccy,  (1) 

we  have  (Art.  309)  x  =  my.  (2) 

Substituting  the  first  pair  of  values, 

12  =  3  m, 
whence  .  m  =  4. 

Substituting  this  and  the  second  given  value  of  x  in  (2), 

32  =  4*/, 
whence  y  =  8. 


192  HIGHER  ALGEBRA 

• 

2i>  Solution.     Since  £xy, 

we  have  (Art.  311)    .  x' :  y'  :  :  a"  :  ?/". 

Substituting  the  given  values, 

12:3::  32  :  y", 
whence  y"  —  8. 

Caution.  In  solving  such  examples,  students  sometimes  sub- 
stitute given  values  directly  in  the  variation,  thus  :  "  12  oc  3,  etc." 
As  12  and  3  cannot  vary,  it  is  absurd  to  write  x  between  them. 

10.  If  xccy,  and  x  =  15  when  y  =  2±,  what  is  the  value  of  x 
when  y  =  4  ? 

11.  If  x  x  -,  and  x  =  10  when  ?/  =  3,  what  is  the  value  of  y 

y 

when  x  =  6  ? 

12.  If  x-  x  — ,  and  x  =  4  when  ?/  =  3,  what  is  the  value  of  x 

y- 

when  y  =  2  ? 

13.  If  a?  x  ?/2,  and  #  =  24  when  y  =  2  and  z  =  3,  what  is  the 
value  of  y  when  a:  =  8  and  z  =  4  ? 

14.  If  x  x  £,  and  a;  =  12  when  y  =  8  and  2  =  2,  what  is  the 

2 

value  of  2  when  x  =  9  and  y  —  12  ? 

15.  If  a?  varies  directly  as  y  and  inversely  as  the  square  of  2, 
and  a;  =  6  when  y  =  8  and  2  =  2,  find  the  value  of  x  when  2/  =  18 
and  2  =  3. 

16.  The  volume  of  a  sphere  varies  as  the  cube  of  its  radius.  If 
the  volume  of  a  soap  bubble  is  5.888  when  its  radius  is  1  inch, 
what  is  its  volume  when  its  radius  is  1\  inches  ? 

Solution.     Since  v  x  r3, 

we  have  (Art.  309)  v  =  mrz. 

Substituting  the  given  values  of  v  and  r, 

m  =  5.888. 
Substituting  this  value  of  to  and  the  second  value  of  r, 
v  =  5.888  (|)3  =  19.872. 


VARIATION  193 

Or,  if  we  choose,  we  may  use  a  proposition,  thus  : 

V  :  v'  : :  r3  :  r'3, 
5.888  :  v'  :  :  1  :  ^, 
whence  V  =  19.872. 

17.  If  a  metal  ball  whose  radius  is  2  inches  weighs  6  pounds, 
what  is  the  weight  of  a  ball  of  the  same  metal  whose  radius  is 
4  inches  ? 

la  The  distance  fallen  by  a  body  from  rest  varies  as  the 
square  of  the  time  of  falling.  If  a  body  falls  257J  feet  in  4  sec- 
onds, how  far  will  it  fall  in  6  seconds  ? 

19.  Amount  of  illumination  varies  directly  as  the  intensity  of 
the  light  and  inversely  as  the  square  of  the  distance  from  the 
light.  What  must  be  the  intensity  of  a  light  to  give  at  the  dis- 
tance of  75  feet  3  times  the  illumination  of  one  whose  intensity 
is  10  and  distance  50  feet  ? 

20.  The  volume  of  a  pyramid  varies  jointly  as  its  base  and 
altitude.  A  pyramid  whose  base  is  9  feet  square  and  whose 
height  is  10  feet  contains  10  cubic  yards.  What  must  be  the 
height  of  a  pyramid  with  a  base  3  feet  square  in  order  that  it 
may  contain  2  cubic  yards  ? 

21.  The  volume  of  a  right  circular  cone  varies  jointly  as  its 
height  and  the  square  of  the  radius  of  its  base.  If  the  volume  of 
a  certain  cone  is  94  cubic  inches,  what  is  the  volume  of  another 
cone  twice  as  high  and  the  radius  of  whose  base  is  half  as  great  ? 

22.  The  volume  of  a  gas  varies  as  the  absolute  temperature 
and  inversely  as  the  pressure.  When  the  temperature  in  a  given 
case  is  260  and  the  pressure  15,  the  volume  is  200  cubic  inches. 
What  will  be  the  volume  when  the  temperature  becomes  390  and 
the  pressure  18  ? 

23.  The  pressure  of  the  wind  on  a  plane  area  varies  jointly  as 
the  area  and  the  square  of  the  velocity  of  the  wind.  If  the 
pressure  on  1  square  foot  is  1  pound  when  the  velocity  of  the 
wind  is  16  miles  an  hour,  what  is  the  velocity  of  the  wind  when 
the  pressure  on  2  square  yards  is  50  pounds  ? 

24.  If  x  -f  y  cc  x  —  y,  prove  that  x2  +  y2  oc  xy. 

downey's  alg.  — 13 


194  HIGHER   ALGEBRA 

25.  Given  y  =p  +  q,  in  which  pccx,  and  q  cc  —     When  x  =  1, 

x 

y  =  6 ;  and  when  x  =  2,  ?/  =  5.     Find  the  value  of  y  in  terms  of  as. 

26.  Given  #  =  a +^:> +</,  in  which  a  is  constant,  p&y,  and 
5  cc  2/2.  When  g  =  6,  17,  34,  ?/  =  1,  2,  3,  respectively.  Find  the 
value  of  a?  in  terms  of  y. 

27.  Given  that  s  oc  t2  when  /  is  constant,  and  8  oc/  when  t  is 
constant;  also  2s  =  /  when  t  =  1.  Find  the  value  of  s  in  terms 
of  /  and  t. 

28.  By  Kepler's  third  law  the  square  of  a  planet's  period  of 
revolution  around  the  sun  varies  as  the  cube  of  its  mean  distance 
from  the  sun.  If  r  is  the  distance  of  a  planet  whose  period  is  t, 
what  is  the  distance  of  a  planet  whose  period  is  t'  ? 

29.  Attraction  varies  directly  as  the  mass  of  the  attracting 
body  and  inversely  as  the  square  of  the  distance  from  its  center. 
The  mass  of  the  sun  being  332,000  times  that  of  the  earth,  and  its 
radius  109.4  times  that  of  the  earth,  what  is  the  weight  of  a 
body  at  the  surface  of  the  sun  as  compared  with  its  weight  at  the 
surface  of  the  earth  ? 

30.  The  mass  of  the  sun  being  1047.35  times  that  of  Jupiter, 
and  the  distance  of  Jupiter  from  the  sun  being  414  times  the 
distance  from  Jupiter  to  his  outer  satellite,  what  is  the  attraction 
of  the  sun  on  Jupiter  as  compared  with  Jupiter's  attraction  on 
his  outer  satellite  ? 


CHAPTER   XIV 

PROGRESSIONS 

SECTION   I  — ARITHMETICAL  PROGRESSION 

312.  A  Series  is  a  succession  of  terms  proceeding  according  to 
a  definite  law. 

313.  A  series  is  Increasing  or  Decreasing,  or  Ascending  or  De- 
scending, according  as  the  terms  increase  or  decrease. 

314.  An  Arithmetical  Progression  is  a  series  in  which  each  term 
is  greater  or  less  than  the  preceding  term  by  a  constant  quantity, 
called  the  Common  Difference. 

If  we  regard  each  term  as  being  obtained  by  the  addition  of 
the  common  difference  to  the  preceding  term,  the  common  differ- 
ence is  plus  in  an  increasing,  and  minus  in  a  decreasing,  arith- 
metical progression. 

Thus,  1,  3,  5,  7,  etc.,  is  an  increasing  arithmetical  progression,  in  which 
the  common  difference  is  2. 

15,  10,  5,  0,  —  5,  —  10,  etc.,  is  a  decreasing  arithmetical  progression,  in 
which  the  common  difference  is  —  5. 

a,  a  +  d,  a  +  2d,  a  +  3d,  etc.,  is  the  general  form  of  an  arith- 
metical progression,  the  common  difference  being  d,  which  may 
be  either  plus  or  minus. 

315.  In  the  treatment  of  arithmetical  progression,  five  ele- 
ments are  involved : 

1.  The  first  term,  a. 

2.  The  last  term,  I. 

3.  The  common  difference,  d. 

4.  The  number  of  terms,  n. 

5.  The  sum  of  the  terms,  S. 

195 


196  HIGHER   ALGEBRA 

316.  Arithmetical  Means  between  any  two  quantities  are  the 
terms  that  lie  between  them  in  an  arithmetical  progression. 

When  there  is  but  one  intermediate  term,  it  is  called  the  Arith- 
metical Mean  of  (or  between)  the  other  two. 

317.  Theorem.  TJie  arithmetical  mean  of  two  quantities  is  equal 
to  half  their  sum. 

Dem.     Let  a,  b,  c  be  in  arithmetical  progression. 

By  definition  b  —  a  =  c  —  b, 

whence  b  =  ^  (a  -f  c). 

318.  Theorem.  TJie  formula  for  the  last,  or  nth,  term  of  an 
arithmetical  progression  in  terms  of  the  first  term,  the  common  dif- 
ference, and  the  number  of  terms,  is 

I  =  a-\-  (n  —  1)  d. 

Dem.     In  the  general  form  of  an  arithmetical  progression, 

a,  a  -f-  d,  a -{-2d,  a  -{-3d,  a  -f  4  d,  etc., 

it  is  seen  that  each  term  is  formed  by  adding  to  a  the  product  of 
d  and  a  number  which  is  1  less  than  the  number  of  the  term; 
that  is, 

I  =  a  -\-  (n  —  l)d. 

319.  Theorem.  The  formula  for  the  sum  of  an  arithmetical  pro- 
gression in  terms  of  the  first  term,  the  last  term,  and  the  number  of 
terms,  is 

S  =  l 


a  +  Z\ 


Dem.  Taking  the  sum  of  the  terms  of  the  general  form  of  an 
arithmetical  progression  and  the  sum  of  the  same  terms  in  the 
reverse  order,  we  have 

#  =  a+0  +  d)  +  («  +  2ff)+  --(l-2d)  +  (l-d)  +  l,       (1) 

and     S  =  I  +  (I  -  d)  +  (I  -  2  d)  +  •  • .  (a  +  2  d)  +  (a  +  d)  +  a.       (2) 

Adding  (1)  and  (2), 

2  S  =  (a  + 1)  +  (a  + 1)  +  (a  + 1)  +  ■  •  •  (a  +  0  +  (a  + 1)  +  (a  + 1). 


ARITHMETICAL   PROGRESSION  197 

It  is  seen  that  (a  4-/)  is  taken  as  many  times  as  there  are 
terms.     Hence 

2  S  =  (a  +  0  n, 


or  S 


m 


320.    Any  one  of  the  four  quantities  involved  in  either  of  the 

equations, 

I  =  a  -f-  (w  —  1)  d, 


£ 


(t> 


may  be  found  in  terms  of  the  other  three ;  and  by  combining  the 
two  equations,  any  one  of  the  five  quantities  involved  may  be 
eliminated,  and  any  one  of  the  remaining  four  found  in  terms  of 
the  other  three.  The  twenty  formulae  on  page  198  result.  They 
are  convenient,  but  not  necessary,  as  all  cases  may  be  solved  by  the 
two  fundamental  formulae,  either  directly  or  by  first  finding  an 
intermediate  element.  The  student  should  reserve,  until  after  he 
has  taken  the  subject  of  Quadratic  Equations,  the  development  of 
numbers  2,  11,  18,  and  20. 

EXAMPLES   LXXIX 

1.  Find  the  20th  term  and  the  sum  of  20  terms  of   1,  4,  7, 
10,  etc. 

2.  Find  the  21st  term  and  the  sum  of  21  terms  of  3,  7,  11, 
15,  etc. 

3.  Find  the  36th  term  and  the  sum  of  36  terms  of  12,  10,  8, 
6,  etc. 

4.  Find  the  10th  term   and   the   sum  of   10  terms   of  3,  2. \, 
If,  etc. 

In  each  of  the  following,  find  the  two  elements  not  given : 
5.   a  =  1,    d  =  5,      n  =  24.  6.    1  =  71,    d  =  5,    n  =  15. 

7.   0=13,  1  =  73,    rc  =  ll.  8.   o  =  10,     1  =  87,  d  =  7. 

9.   d  =  4,    n  =  U,     s  =  812.  10.   a=  -5,  ?i  =  19,  s=-950. 

ll.a  =  7,     1  =  143,  s  =  1350.        12.    l=-§,  n  =  19,  8  =  0. 


198  HIGHER   ALGEBRA 

Formula  in  Arithmetical  Progression 


Number 


Given 


Required 


Formulae 


4. 


a,  d,  n 

a,  d,  S 

a,  7i,  S 

d,  n,  S 


a,   d,    I 

a,   n,    I 
d,   7i,    I 


l=za  +  (n  —  l)cl, 

I  =  -  ^d  ±V \2  dS  +(a  -  \df\ 


2S 


a, 


1  = 


S      (n-l)d 
n  2 


S 


\n\2a+(n-l)d\, 

2  2d  ' 

(*  +  «)§ 
\n\2l-(n-l)d\. 


9. 
10. 

11. 

12. 

13. 
14. 
15. 
16. 


d,  7i,  I 

d,  n,  S 

d,  I,  S 

n,  I,  .S 


a,  71,  I 

a,  n,  S 

a,  /,  S 

n,  I,  S 


l-(7i-l)d, 

S      (n  -  l)d 
n  2       ' 


=  i  d  ±  V(7  +  i  df  -  2  dS, 


I  —  a 

w  (w  —  1) 
I2 -a2 
2S-l-a 
2(nl-S) 

7%  (n  —  1) 


17. 
18. 
19. 
20. 


a,   d,    I 


a,  d,  S 
a,  I,  S 
d,    I,    8 


d 


+  1, 


±  V(2  a-ri)'+8  dS-2  q+rf 
2d 

=  2S  ; 

=  2  Z  +  ^  ±  V(2  I  +  d)2  -^WdS 
2d 


ARITHMETICAL   PROGRESSION  199 

13.  Insert  3  arithmetic  means  between  73  and  193. 
Sue     First  find  <?,  n  being  5. 

14.  Insert  11  arithmetic  means  between  —  49  and  59. 

15.  Find  the  first  term  of  an  arithmetical  progression  whose 
59th  term  is  —  2\,  and  60th  term  —  If. 

16.  Find  the  first  term  of  an  arithmetical  progression  whose 
2d  term  is  \,  and  55th  term  5.8. 

17.  Find  the  sum  of  24  terms  of  an  arithmetical  progression 
whose  13th  term  is  25,  and  19th  term  37. 

18.  Find  the  sum  of  12  terms  of  the  arithmetical  progression 
whose  1st  term  is  38,  and  4th  term  86. 

19.  Between  what  two  terms  of  3,  8,  13,  etc.,  does  391  lie  ? 

20.  If  a  body  falling  from  rest  descends  a  feet  the  first  second, 
3  a  the  second,  5  a  the  third,  and  so  on,  how  far  will  it  fall  during 
the  tth  second,  and  how  far  in  t  seconds  ? 

21.  If  a  body  falling  from  rest  descends  16^  feet  the  first 
second,  48 J;  feet  the  second,  80T%  feet  the  third,  and  so  on,  how 
far  will  it  fall  in  30  seconds  ? 

22.  A  man  training  for  a  mile  race  ran  over  the  course  every 
day  for  21  days.  His  time  the  first  day  was  8  minutes^  but  this 
was  gradually  diminished,  his  whole  time  for  the  21  runs  being 
133  minutes.  What  was  the  average  daily  diminution  of  time, 
and  what  was  his  time  the  last  day  ? 

23.  Some  fence  posts  are  to  be  carried  from  a  pile  to  the  holes 
where  they  are  to  be  set.  How  far  must  a  laborer  walk,  carrying 
one  post  at  a  time,  to  place  them  in  the  50  post  holes,  which  are 
in  a  straight  line  and  8  feet  apart,  the  first  one  being  at  the  pile  ? 

24.  A  man  bought  an  estate  which  yielded  $  1500  profit  the 
first  year.  His  personal  expenses  for  the  first  year  were  $  1250. 
His  income  from  the  estate  increased  $  100  yearly,  and  his 
personal  expenses  increased  $  125  yearly.  After  how  many 
years  were  his  personal  expenses  equal  to  the  income  from  his 
estate  ? 


200  HIGHER  ALGEBRA 

SECTION   II.  — GEOMETRICAL   PROGRESSION 

321.  A  Geometrical  Progression  is  a  series  in  which  each  term 
is  eqnal  to  the  preceding  term  multiplied  by  a  constant  quantity, 
called  the  Ratio. 

The  ratio  is  greater  than  unity  in  an  increasing  and  less  than 
unity  in  a  decreasing  geometrical  progression. 

Thus,  2,  4,  8,  10,  etc.,  is  an  increasing  geometrical  progression  in  which 
the  ratio  is  2. 

12,  4,  |,  |,  547,  etc.,  is  a  decreasing  geometrical  progression  in  which  the 
ratio  is  ^. 

a,  ar,  ar2,  ar3,  arA,  etc.,  is  the  general  form  of  a  geometrical 
progression,  the  ratio  being  r. 

322.  In  the  treatment  of  geometrical  progression,  five  elements 
are  involved: 

1.  The  first  term,  a. 

2.  The  last  term,  I. 

3.  The  ratio,  r. 

4.  The  number  of  terms,  n. 

5.  The  sum  of  the  terms,  S. 

323.  Geometric  Means  between  any  two  quantities  are  the  terms 
that  lie  between  them  in  a  geometrical  progression. 

When  there  is  but  one  intermediate  term,  it  is  called  the 
Geometric  Mean  of  (or  between)  the  other  two. 

324.  Theorem.  The  geometric  mean  of  two  quantities  is  equal  to 
the  square  root  of  their  product. 

Dem.     Let  a,  b,  c,  be  in  geometric  progression. 

By  definition  -  =  -, 

a      b 

whence  b  =  V«c. 

325.  Sch.  The  geometric  mean  of  two  quantities  is  the  same 
as  the  mean  proportional  between  them  (Arts.  287  and  295). 


GEOMETRICAL  PROGRESSION  201 

326.  Theorem.  The  formula  for  the  last,  or  nth,  term  of  a 
geometrical  progression  in  terms  of  the  first  term,  the  ratio,  and  the 
number  of  terms,  is 

l  =  arn~\ 

Dem.     In  the  general  form  of  a  geometrical  progression, 

a,  ar,  ar2,  ar3,  ar*,  etc., 

it  is  seen  that  each  term  is  formed  by  multiplying  a  by  r  affected 
with  an  exponent  1  less  than  the  number  of  the  term ;  that  is, 

I  =  arn~l. 

327.  Theorem.  The  formula  for  the  sum  of  a  geometrical  pro- 
gression in  terms  of  the  first  term,  the  ratio,  and  the  number  of 
terms,  is 

arn  —  a 


S  = 


r-1 


Dem.  Taking  the  sum  of  the  terms  of  the  general  form  of  a 
geometrical  progression  and  multiplying  this  series  by  r,  we 
have 

S  =  a  +  ar  +  ar2  +  •••  arn~3  +  arn~2  +  arn~l,  (1) 

and  rS  =         ar  -f  ar  +  *..«**-*  4.  arn  - 2  +  arn  ~ 1  +  arn.    (2) 

Subtracting  (1)  from  (2), 

rS  —  S  =  arn  —  a, 

whence  ^  =  ar"~a. 

r  —  1 

328.  Theorem.  The  formula  for  the  sum  of  a  geometrical  pro- 
gression in  terms  of  the  first  term,  the  last  term,  and  the  ratio,  is 

a     Ir  —  a 

Dem.     From  l  =  arn~1 

we  have  lr  =  aiM. 

Substituting  this  value  of  arn  in  the  formula  for  S,  Art.  327, 
c     Ir  —  a 


202  HIGHER   ALGEBRA 

Formula  in  Geometrical  Progression 


Number  |       Given 


a,  r,  n 

a,  r,  jS 

a,  n,  S 

r,  n,  S 


Required 


FOHMULuE 


I  =  at*"1, 

r 
l(S  -  l)n~l  -a(S~  ay-1  =  0, 
l==(r-  l)^"-1 
rn  —  1 


a.  r,  n 
a,  r,  I 
a,    n,    I 


S 


s  = 
s  = 
s  = 
s  = 


r  —  1 
rl  —  a 


r-1 

n—l /IT,        n~  1/~^i 

v  <   —    v  a 

n— 1/7         B— 1/~~ 

v  t  —  v  a 


10. 

11. 
12. 


13. 
14. 
15. 
16. 


17. 
18. 
19. 
20. 


r,  n,  I 

r,  n,  S 

r,  I,  S 

n.  I,  S 


a,  n,  I 

a,  n,  S 

a,  I,  S 

n,  I,  S 


a,    r,   S 

a,    I,    S 
r,    h    S 


a  =  — ? 

rn  —  l 


a  =  rl-(r-l)S, 

a(S-a)n  1-l(S-l)n-1  =  0. 


rn r  H =  0, 

a  a 


r  = 


r«_ 


£-/ 


-f 


^  —  z 


0. 


log? -log  a      -j 
logr 
_  log  [a  +  (r  -  1)3]  -  log  a 
log  r 
log  Z  —  log  a 


\og(S-a)-\og(S-l) 


+  1, 


_logZ-log[/r-(r-l)£]  |  L 


logr 


GEOMETRICAL   PROGRESSION  203 

329.    Theorem.      The  formula  for  the   sum   of  an   infinite,  de- 
creasing geometrical  x>rogression  is 

s=    a 


1-r 


Of" 

—  a 

r  - 

-1  ' 

Ir- 

-a 

Dem.     Since  in  a  decreasing  geometrical  progression  r  is  less 
than  unity,  the  term  arn,  in  the  formula 

c      arn  —  a 

*>  = 7~> 

r—  1 

has  no  appreciable  value  when  n  is  infinite.  Hence  this  formula 
becomes 

o_  —  a  _     a 
~r-l~r~r' 

330.    Any  one  of  the  four  quantities  involved  in  any  one  of  the 
equations, 

I  =  arn  " l, 

JS  = 

s  = 

r  —  1 

may  be  found  in  terms  of  the  other  three ;  and  by  combining  the 
first  with  one  of  the  others,  any  one  of  the  five  quantities  in- 
volved may  be  eliminated,  and  any  one  of  the  remaining  four 
found  in  terms  of  the  other  three,  except  in  a  few  cases  involving 
higher  literal  equations.  These  last  can  be  solved  when  numeri- 
cal quantities  are  substituted  for  the  literal.  The  twenty  formulae 
on  page  202  result.  The  student  should  reserve  until  after  taking 
the  subject  of  Logarithms  the  development  of  the  last  four. 

EXAMPLES   LXXX 

1.   Prove  the  following  properties  of  a  geometrical  progression : 

(a)  The  alternate  terms,  or  any  terms  separated  by  the  same 
intervals,  are  in  geometrical  progression. 

(b)  The  products  of  the  terms  by  the  same  quantity  are  in 
geometrical  progression. 


204  HIGHER  ALGEBRA 

(c)  The  same  powers  of  the  terms  are  in  geometrical  pro- 
gression. 

(d)  The  reciprocals  of  the  terms  are  in  geometrical  progression. 

(e)  Any  term  is  a  mean  proportional  between  any  two  terms 
separated  from  it  by  the  same  intervals. 

(/)  The  product  of  any  odd  number,  p,  of  consecutive  terms 
is  equal  to  the  pth  power  of  the  middle  one. 

(g)  The  product  of  any  two  terms  is  equal  to  the  product  of 
any  other  two  terms  separated  from  these  by  the  same  intervals 
in  opposite  directions. 

2.  Find  the  12th  term  and  the  sum  of  12  terms  of  1,  2,  4,  8,  etc. 

3.  Find  the  10th  term  and  the  sum  of  10  terms  of  1, 3,  9,  27,  etc. 

4.  Find  the  11th  term  and  the  sum  of  11  terms  of  ^,  Jg,  J,  1, 
etc. 

5.  Find  the  11th  term  and  the  sum  of  11  terms  of  3,  —  6,  12, 
-  24,  etc. 

In  each  of  the  following  find  the  two  elements  not  given : 

6.  r  =  2,      w  =  8,    £  =  1275.      7.    r  =  3,      n  =  9,    I  =  26,244. 
8.   a  =  -  |,  n  =  7,     r  =  -  J.        9.    I  =  256,  n  =  9,  r  =  2. 

10.   a  =  -2,?i  =  6,     Z  =  2048.    11.   a  =  2,      n  =  7,    I  =  145. 
12.  a  =  1,      I  =  81,  r  =  3.         13.    Z  =  160,  r  =  2,  S  =  315. 

14.  Insert  3  geometric  means  between  17  and  4352. 
Sug.     First  find  ?%  n  being  5. 

15.  Insert  4  geometric  means  between  26  and  6318. 

16.  Find  the  1st  term  of  a  geometrical  progression  whose  5th 
term  is  336,  and  9th  term  5376. 

17.  Find  the  sum  of  8  terms  of  the  geometrical  progression 
whose  4th  term  is  108,  and  7th  term  2916. 

18.  Find  the  11th  term  of  a  geometrical  progression  whose  7th 
term  is  192,  and  10th  term  -  1536. 

19.  The  sum  of  the  first  8  terms  of  a  geometrical  progression 
is  17  times  the  sum  of  the  first  4  terms.     Find  the  ratio. 


HARMONIC  PROGRESSION  205 

20.  The  1st  term  of  a  geometrical  progression  is  3,  and  the 
sum  of  the  first  3  terms  is  one  eighth  of  the  sum  of  the  next 
3  terms.     Find  the  ratio. 

21.  The  sum  of  the  1st  and  2d  terms  of  a  geometrical  progres- 
sion is  30,  and  the  sum  of  the  4th  and  5th  is  1920.  Find  the  first 
5  terms  of  the  progression. 

22.  Find  the  sum  of  1,  £,  \,  |,  etc.,  to  infinity. 

23.  Find  the  value  of  1  -f-  \  +  i  -f  ^7  +  etc.,  to  infinity. 

24.  Find  the  value  ofl  —  ^  +  i  —  2V  +  etc.,  to  infinity. 

25.  Find  the  value  of  .423. 

Sug.     .423  =  .4232323  .  .  .  =.4  +  .023  +  .00023  +  etc. 

26.  Find  the  value  of  .27. 

27.  Find  the  value  of  .3i2. 

28.  Each  term  in  a  certain  infinite  decreasing  geometrical  pro- 
gression is  equal  to  the  sum  of  all  that  follow  it.     Find  the  ratio. 

29.  What  is  the  distance  passed  through  before  coming  to  rest 
by  a  ball  which  falls  from  a  height  of  50  feet  and  at  every  fall 
rebounds  half  the  distance  ? 

30.  A  "letter  chain"  is  started  for  the  benefit  of  a  public 
charity,  three  letters,  each  numbered  1,  being  sent  out  by  the 
starter  with  the  request  that  each  of  the  recipients  remit  10  cents 
and  send  out  three  other  letters,  each  numbered  2,  with  a  similar 
request,  and  so  on,  until  the  numbers  reach  25.  Should  all  com- 
ply, («)  How  much  would  be  realized  for  the  charity  ?  (b)  What 
would  be  the  entire  outlay  for  postage  at  2  cents  for  each  letter  ? 
(c)  With  a  uniform  distribution,  how  many  times  would  each  of 
the  75  million  inhabitants  of  the  United  States  respond  ? 

SECTION   III.  — HARMONIC   PROGRESSION 

331.  A  Harmonic  Progression  is  a  series  of  terms  whose  recipro- 
cals are  in  arithmetical  progression. 

Thus,  1,  \,  f,  ^,  ^5,  etc.,  are  in  harmonic  progression,  because  their 
reciprocals,  1,  4,  7,  10,  13,  etc.,  are  in  arithmetical  progression. 


206  HIGHER  ALGEBRA 

The  general  form  of  a  harmonic  progression  is  -, ,  — , 

-^  a  a  +  d   a  +  2d 

etc.,  in  which  a  is  the  first  term  of  an  arithmetical  pro- 


a  +  3d 

gression,  and  d  the  common  difference. 

Most  problems  concerning  quantities  in  harmonic  progression 
are  solved  by  treating  the  arithmetical  progression  obtained  by 
taking  the  reciprocals  of  the  quantities  that  are  in  harmonic 
progression. 

332.  If  the  lengths  of  strings  of  the  same  substance,  size,  and 
tension  be  proportional  to  the  terms  of  a  harmonic  progression, 
any  two  of  these  strings  vibrating  together  produce  harmony  of 
sound ;  hence  the  term  harmonic. 

333.  Harmonic  Means  between  any  two  quantities  are  the  terms 
that  lie  between  those  quantities  in  a  harmonic  progression. 

When  there  is  but  one  intermediate  term,  it  is  called  the  Har- 
monic Mean  of  (or  between)  the  other  two. 

334.  Theorem.  The  harmonic  mean  of  two  quantities  is  twice 
their  product  divided  by  their  sum. 

Dem.     If  a,  b,  c  are  in  harmonic  progression,  -,  -,  -  are,  by 

a    b    c 
definition,  in  arithmetical  progression ;  hence 

11^11 

b      a     c      b 

whence  b  =  — '■ — 

a  +  c 

335.  Theorem.  If  three  quantities  are  in  harmonic  progression, 
the  difference  between  the  first  and  second  is  to  the  difference  between 
the  second  and  third  as  the  first  is  to  the  third. 

Dem.     If  a,  b,  c  are  in  harmonic  progression,  -,  ->  -  are,  by 

a   b   c 
definition,  in  arithmetical  progression ;  hence 

1_1=1_1 

b      a     c      b 
Clearing  of  fractions,  ac  —  bc  =  ab  —  ac, 

or  c(a  —  b)  =  a(b  —  c)  ; 

therefore  (Art.  297)        a  —  b  :  b  —  c  :  :  a  :  c. 


HARMONIC  PROGRESSION  207 

336.  Theorem.  The  formula  for  the  last,  or  nth,  term  of  a  har- 
monic progression  in  terms  of  the  first  term,  the  second  term,  and  the 
number  of  terms,  is 

ab 


I 


b  +  (n—l)(a  —  b) 


Dem.     If  a,  b,  etc.,  are  in  harmonic   progression,  -,  -,  etc.,  are, 

d   b 
by  definition,  in  arithmetical  progression ;  hence 

b     a        ab 
Substituting  in  V  =  a'  +  (»  -  l)d  (Art.  318), 

1-1  , T  (n      !)<*  -  o -  o  +(*  -  i)(o  -  b) 

la  ab  ab 

whence  I  = 


b  +  {n  -  1)  (a  -  b) 

337.   Note.    No  formula  for  the  sum  of  a  harmonic  progression 
is  known. 

EXAMPLES  LXXXI 

1.  Find  the  23d  term  of  J,  \,  y1^,  y1^,  etc. 

2.  Find   the  4th   and  -7th   terms   of   a  harmonic   progression 
whose  2d  term  is  \,  and  5th  term  y1^. 

3.  Find  the  harmonic  mean  of  21  and  42. 

4.  Insert  3  harmonic  means  between  14  and  42. 

Sue     Find  3 '  arithmetic   means  between   Ta¥  and   fa   and   take   their 
reciprocals. 

5.  Insert  3  harmonic  means  between  -J^  and  fa. 

6.  If  a,  b,  c,  d  are  in  harmonic  progression,  show  that 

ab  :  cd  : :  a  —  b  :  c  —  d. 

7.  Show  that  the  geometric  mean  of  two  numbers  is  also  the 
geometric  mean  of  their  arithmetic  and  harmonic  means. 


CHAPTER   XV 
QUADRATIC  EQUATIONS 

338.  Quadratic  Equations  (Art.  238)  are  distinguished  as  Pure 
(called  also  Incomplete)  and  Affected  (called  also  Complete). 

339.  A  Pure  Quadratic  Equation  is  an  equation  which  contains 
no  power  of  the  unknown  quantity  but  the  second. 

Thus,  ax2  +  b  =  cd  and  3  x2  =  108  are  pure  quadratic  equations,  or  pure 
quadratics. 

340.  An  Affected  Quadratic  Equation  is  an  equation  which  con- 
tains both  the  first  and  the  second  powers  of  the  unknown 
quantity. 

Thus,  x2  —  4  x  =  12,  2  x2  +  7  x  -  18  =  0,*  ax2  +  bx  =  c,  are  affected  quad- 
ratic equations,  or  affected  quadratics. 

341.  A  Root  of  an  Equation  is  a  quantity  which,  substituted  for 
the  unknown  quantity,  satisfies  the  equation. 

PURE   QUADRATICS 

342.  Prob.     To  solve  a  pure  quadratic  equation. 

Rule.  By  clearing  of  fractions,  transposing,  uniting  terms,  and 
dividing  by  the  coefficient  of  the  square  of  the  unknown  quantity, 
reduce  the  equation  to  the  form  x1  =  m ;  then  extract  the  square  root 
of  both  members,  giving  the  double  sign  to  the  second  member  of  the 
residt. 

Dem.  All  of  the  operations  of  reducing  to  the  form  x2  =  m, 
and  the  extraction  of  the  square  root  as  well,  affect  both  mem- 
bers alike,  and,  consequently  (Art.  243),  do  not  destroy  the 
equality  of  the  members.  As  these  operations  leave  in  the  first 
member  simply  the  unknown  quantity,  the  equation  is  solved. 

208 


QU  ABU  AT  [C  EQUATIONS  209 

That  the  result  should  have  the  double  sign  is  evident  from 
the  fact  that  a  quantity  has  two  square  roots  numerically  equal, 
but  with  opposite  signs. 

343.  Cor.  The  roots  of  a  pure  quadratic  equation  are  both  rational 
or  both  surd,  both  real  or  both  imaginary. 

EXAMPLES   LXXXn 
Solve  the  following : 

1.   11  a,-2 -44  =  5^ +  10.  2.   5ar2-9  =  2a,-2-f-G6. 

5         8         5 


3.   0  +  2)2  =  4z  +  5. 


4  —  x     3     4  +  x 


5.   _i--  +  —5—-X25.  6.   2(x+3)(x-3)  =  (x+iy-2x. 

J.  —  L  X        L  -j-  —  X 


7.   x2  —  ax  -j-  b  =  ax  (x  —  1).  8.   xy/a?  -+-  (J  =  a;2  4- 1. 


rt    «  .  V«2  —  a,*2     ar  _•  .     /  2  ,     n  2  a2 

9.   -  +  — =  —  10.    x  H-  VaJ  4-  a;2 


x  x  b  Va2  +  a^ 


11  2  +  2  =a?.     12.    ^  +  l+VaVZLl=;to 


PROBLEMS   LEADING  TO   PURE   QUADRATIC   EQUATIONS 
EXAMPLES  LXXXHI 

1.  Find  two  numbers  in  the  ratio  of  2  to  5,  the  sum  of  whose 
squares  is  261. 

2.  Find  three  numbers  which  shall  be  to  one  another  as  m,  n, 
and  p,  and  the  sum  of  whose  squares  shall  be  $. 

3.  Divide  21  into  two  such  parts  that  the  square  of  the  less 
shall  be  to  that  of  the  greater  as  4  to  25. 

4.  The  sides  of  two  square  rooms  are  in  the  ratio  of  2  to  3,. 
and  the  larger  room  requires  20  square  yards  more  of  carpet  than 
the  smaller.     Find  a  side  of  each. 

5.  Two  square  plats  of  ground  contain  272  square  rods,  and 
a  side  of  the  larger  is  as  much  greater  than  10  rods  as  a  side  of 
the  other  is  less  than  10  rods.     Find  a  side  of  each. 

Downey's  alg. — 14 


210  HIGHER  ALGEBRA 

6.  A  rectangular  field  whose  length  is  li  times  its  breadth 
contains  9  acres.     Find,  the  length,  in  rods,  of  each  side. 

7.  An  army  was  formed,  with  5  more  men  in  file  than  in  rank ; 
but  when  the  form  was  changed  so  that  there  were  845  more  men 
in  rank  than  before,  there  were  but  5  ranks.  Find  the  number 
of  men  in  the  army. 

8.  A  boat's  crew  can  row  in  still  water  at  the  rate  of  9  miles 
an  hour.  If  it  take  the  crew  2\  hours  to  row  9  miles  up  a  river 
and  back  to  the  starting  point,  what  is  the  rate  of  the  current  ? 

9.  The  distances  through  which  a  body  falls  being  as  the 
squares  of  the  times,  and  the  distance  fallen  during  the  first 
second  being  lGy1^  feet,  in  what  time  will  a  body  fall  500  feet  ? 
In  what  time  will  it  fall  a  mile  ? 

10.  The  mass  of  the  earth  is  332,000  times  that  of  the  earth, 
and  the  distance  between  the  two  bodies  is  93,000,000  miles. 
How  far  from  the  earth,  between  the  earth  and  the  sun,  is  the 
point  of  equal  attraction,  the  law  of  attraction  being  that  it 
varies  directly  as  the  mass  and  inversely  as  the  square  of  the 
distance  ? 

Sue  In  this  and  some  of  the  following  problems,  to  avoid  an  affected 
quadratic  equation,  take  the  square  root  before  performing  indicated  opera- 
tions. 

11.  The  intensities  of  two  lights  are  as  7  :  17,  and  their  dis- 
tance apart  132  feet.  Where  in  the  line  of  the  lights  are  the 
points  of  equal  illumination,  assuming  that  the  amounts  of 
illumination  are  to  each  other  directly  as  the  intensities  and 
inversely  as  the  squares  of  the  distances  ? 

12.  The  loudness  of  one  church  bell  is  three  times  that  of 
another.  If  the  amount  of  sound  varies  directly  as  the  loudness 
and  inversely  as  the  square  of  the  distance,  where  on  the  line  of 
the  two  will  the  bells  be  equally  well  heard,. the  distance  between 
them  being  a  ? 

13.  A  girl  worked  two  square  pieces  of  worsted  work  of  the 
same  kind,  the  edges  of  one  being  an  inch  longer  than  those  of 
the  other ;  one  took  12^  skeins  and  the  other  18  skeins.  How 
long  were  the  edges  of  the  smaller  piece  ? 


QUADRATIC  EQUATIONS  211 

14.  From  two  towns  m  miles  apart,  two  persons,  A  and  B, 
started  at  the  same  time  and  traveled  toward  each  other.  When 
they  met,  A,  the  faster  traveler,  had  gone  n  miles,  the  time  on 
the  road  being  equal  to  the  difference  of  their  rates.  Find  their 
rates. 

15.  A  and  B  are  two  stations  300  miles  apart.  Two  trains 
start  simultaneously  from  A  and  B,  each  to  the  opposite  station. 
The  train  from  A  reaches  B  9  hours,  and  the  train  from  B  reaches 
A  1  hours,  after  they  meet.     Find  the  rate  of  each  train. 

16.  Two  travelers,  A  and  B,  started  at  the  same  time  from 
two  different  places,  C  and  D  respectively,  and  traveled  toward 
each  other.  When  they  met,  it  appeared  that  A  had  gone  30  miles 
more  than  B ;  also  that  A  could  reach  D  in  4  days,  and  B  could 
reach  C  in  9  days.     Find  the  distance  from  C  to  D. 

17.  Two  bicyclists  start  at  the  intersection  of  two  roads  at 
right  angles  to  each  other  and  ride,  one  on  each  road,  at  rates  of 
12  and  16  miles  an  hour  respectively.  In  how  many  minutes 
will  they  be  8  miles  apart  ? 

AFFECTED  QUADRATICS 
344.  Prob.  To  solve  an  affected  quadratic  equation. 
Rule.  Reduce  to  the  form  x2  +px  =  q  (in  ivhich  p  and  q  may  be 
positive  or  negative,  integral  or  fractional) ;  then  the  roots  are  half 
of  the  coefficient  of  the  second  term  taken  with  the  opposite  sign,  ± 
the  square  root  of  the  sum  of  the  square  of  this  half  coefficient  and 
the  absolute  term. 

Dem.     All  of  the  operations  of  reducing  to  the  form 

x*  +px  =  q, 

viz.,  clearing  of  fractions,  transposing,  uniting  terms,  and  divid- 
ing by  the  coefficient  of  the  second  power  of  the  unknown  quantity, 
affect  both  members  alike,  and,  consequently  (Art.  243),  do  not 
destroy  the  equality  of  the  members. 

P2 
Now  if  -j  be  added  to  both  members  of  the  equation,  giving 


212  HIGHER  ALGEBRA 

the  equality  will  not  be  destroyed  and  the  first  member  will  be  a 
perfect  square,  since  the  middle  term  is  twice  the  product  of  the 
square  roots  of  the  other  two  (Arts.  61  and  62).  This  operation  is 
called  completing  the  square.  Extracting  the  square  root  of  both 
members,  which  does  not  destroy  the  equality,  we  have 


I=±v? 


tflF+ffi 


or 


which  corresponds  with  the  statement  in  the  rule. 

345.  Cor.  i.  If  the  first  term  of  the  roots  is  numerically  greater 
than  the  radical  term,  both  roots  have  the  sign  of  the  first  term;  if 
numerically  less,  one  root  is  plus  and  one  minus.     If  q  is  negative 

and  numerically  greater  than  £-,  both  roots  are  imaginary.     If  q  is 

p2 
negative  and  numerically  equal  to  -j,  the  two  roots  are  equal. 

346.  Cor.  2.     The  sum  of  the  two  roots  is  —  p  and  the  product  is 

-q- 

EXAMPLES  LXXXIV 
Solve  the  following : 

1.   Sx2  -29  =  18  a? -8. 

Solution.  Transposing,  uniting  terms,  and  dividing  by  the  coefficient  of 
x2,  this  becomes 

x*-Qx  =  7. 

The  student  should  not  complete  the  square  —  that  has  been  done  once  for 
all  in  the  demonstration  of  the  rule  —  but  should  write  the  roots  at  once  from 
the  formula  obtained  by  solving  the  general  case.  The  roots  being  half  of  the 
coefficient  of  the  second  term  taken  with  the  opposite  sign,  ±  the  square  root 
of  the  sum  of  the  square  of  this  half  coefficient  and  the  absolute  term,  we 
have  from 

x2  -  6  x  =  7, 

ac  =  8±4  =  7or-l. 

The  full  form  is 

x  =  3  ±  VPT^  =3±4  =  7or-l; 


QUADRATIC  EQUATIONS  213 

but  when  the  numbers  are  small,   the  operations  are  readily  performed 
mentally. 

2.  15-5^=12^. 


Solution.     Reduced  to  the  required  form,  this  becomes 

x2  +  5  x  =  0, 

henc 

e 

»  =  -f  ±1  =  1  or  -6. 

3. 

x2  —  4  x  =  5. 

4.    a;2  +  6  a?  =  -  5.          5.   x2  -  10  or  =  11. 

6. 

x2  +  8  x  =  9. 

7.  >—  12  *  =—11.     8.   x2  -  2  x  =  48. 

9. 

x2  —  6  x  =  16. 

10.   3  aV-  +  36  =  24  x.     11.   a;2  —  4  x  =  60. 

12. 

p*  _  7  a  =  8. 

13.    ar2  -  30  x  =  64.        14.   ar  -  12  a:  =  28. 

15. 

x2  -  12  a  =  45. 

16.   x2  -  8  x  =  33.          17.   6  a,-2  -  21  a;  =  12. 

18         *      -§ 

a?  —  1              g    a¥     2  aa;     62     ^ 

x-1      2 

o0      *  +  2- 

8                      -1    x2  \  a2~b2x-l 

5        3 

a;  -f-  4                                     «6 

/72 

22     ra*2  -4-  — 

*a_oax         o3    **»-«&_» 

c 

_za*.        23.     3ft_2a;  _4 

PROBLEMS  LEADING  TO   AFFECTED   QUADRATIC   EQUATIONS 

347.  When  a  problem  gives  rise  to  a  quadratic  equation,  the 
student  should  note  whether  both  values  obtained  by  the  solution 
of  the  equation  are  admissible  and  capable  of  interpretation.  A 
problem  often  contains  restrictions,  expressed  or  implied,  which 
cannot  be  incorporated  in  the  equation,  and  hence  the  equation 
may  contain  a  root  that  will  not  conform  to  this  restriction.  For 
example,  in  a  problem  involving  as  the  unknown  quantity  the 
digits  of  a  number,  sheep,  cattle,  men,  etc.,  any  fractional  root 
must  be  rejected.  In  some  problems  negative  roots  must  be 
rejected.  Imaginary  roots  indicate  incompatibility  of  conditions 
in  the  problem  (Art.  227). 

EXAMPLES  LXXXV 
1.  Jn  a  number  consisting  of  two  digits  the  tens'  digit  is  1  less 
than  the  square  of  the  units'  digit,  and  when  45  is  subtracted  from 
the  number,  the  digits  change  places.     Find  the  number. 


214  HIGHER  ALGEBRA 

Solution.  Let  x  =  the  units'  digit ; 

then  (x2  —  1)  =  the  tens'  digit, 

and  we  have  from  the  conditions, 

10(x2  -  1)  +  x  -  45  =  10 x  +  x2  -  1, 

or  x2  —  x  =  6, 

whence  x  =  J  ±  f  =  8  or  —  2. 

The  root  —  2  must  be  rejected,  as  the  digits  of  a  number  must  be  positive. 
Hence  the  units'  digit  is  3,  the  tens'  digit  is  32  —  1  =  8,  and  the  number  is  83. 

2.  Two  boys  are  coaching  for  an  examination  in  Greek.  One  of 
them  reads  a  page  the  first  day,  and  each  succeeding  day  reads 
one  more  page  than  the  day  before.  The  other,  beginning  5  days 
later,  reads  12  pages  a  day.  In  how  many  days  will  they  have 
read  the  same  number  of  pages  ? 

Solution.     Let  x  =  the  first  boy's  time  ; 
then  "    x  —  5  =  the  second  boy's  time, 

and  (x  —  5)12  =  the  number  of  .pages  read  by  the  second  boy. 

The  number  of  pages  read  by  the  first  boy  is  the  sum  of  an  arithmetical 
progression  in  which  the  first  term  is  1,  the  number  of  terms  x,  and  the  last 
term  x;  i.e.  (Art.  319), 

(^)»=(Lf> 

Hence 

or 

and 

Both  roots  of  the  equation  are  admissible.  The  second  boy  overtakes  the 
first  in  8  days  and  is  overtaken  by  him  in  15  days  (from  the  beginning),  since 
the  first  is  reading  at  an  increasing  rate. 

3.  Divide  48  into  two  such  parts  that  their  product  may  be  432. 

4.  Divide  24  into  two  such  parts  that  their  product  may  be  35 
times  their  difference. 

5.  For  a  journey  of  108  miles,  G  hours  less  would  have  sufficed, 
had  the  traveler  gone  3  miles  an  hour  faster.  At  what  rate  did 
he  travel  ? 


c 

**)■- 

(x  -  5)  12, 

x- 

!  -  23  x  = 

:    -  120, 

X  = 

.  23      7  _ 
"  2       2 

:  15  or  8. 

QUADRATIC    EQUATIONS  215 

6.  The  length  of  a  rectangle  is  10  feet  more  than  the  breadth, 
and  the  area  is  600  square  feet.  Find  the  length  and  breadth  of 
the  rectangle. 

7.  If  a  bar  of  iron  weighing  60  pounds  be  drawn  out  3  feet 
longer,  it  will  weigh  1  pound  less  per  linear  foot.     Find  its  length. 

8.  A  man  bought  two  farms  for  $  2800  each.  The  larger  con- 
tained 10  acres  more  than  the  smaller,  but  he  paid  $  5  more  per 
acre  for  the  smaller  than  for  the  larger.  How  many  acres  did 
each  contain  ? 

9.  A  man  paid  $  300  for  a  drove  of  sheep.  By  selling  all  but 
10  of  them  at  a  profit  of  $  2.50  each,  he  received  the  amount  he 
paid  for  all  the  sheep.     How  many  sheep  did  he  buy  ? 

10.  It  takes  a  boat's  crew  4  hours  and  12  minutes  to  row  12  miles 
down  a  river  and  back.  If  the  rate  of  the  current  is  3  miles  an 
hour,  at  what  rate  can  the  crew  row  in  still  water  ? 

11.  Two  steamers  ply  between  the  same  two  ports  a  distance 
of  420  miles.  One  goes  £  mile  per  hour  faster  than  the  other, 
and  is  2  hours  less  on  the  voyage.     At  what  rates  do  they  go  ? 

12.  The  plate  of  a  mirror,  18  inches  by  12,  is  to  be  surrounded 
by  a  plain  frame  whose  surface  shall  be  equal  to  that  of  the 
glass.     Find  the  width  of  the  frame. 

13.  A  man  bought  some  sheep  for  $  360,  and  his  neighbor 
bought  6  more  for  the  same  amount,  paying  $  5  less  for  each. 
How  many  did  the  first  man  buy,  and  what  was  the  price  of  each? 

14.  A  man  bought  shares  in  a  company  for  $  375.  A  later 
investor,  after  the  shares  had  declined  $  6.25  each,  bought  for 
the  same  amount  five  more  than  did  the  first  man.  How  many 
shares  did  the  first  man  buy  ? 

15.  A  battalion  of  soldiers,  when  formed  into  a  solid  square, 
presents  16  men  fewer  in  the  front  than  when  formed  into  a 
hollow  square  four  deep.     Required  the  number  of  men. 

16.  Two  vessels,  one  of  which  sails  faster  than  the  other  by  2 
miles  an  hour,  start  together  for  different  ports.  The  faster  ves- 
sel completes  its  voyage  of  1152  miles  1  day  later  than  the  other 
completes  its  voyage  of  720  miles.     What  is  the  rate  of  the  faster 

vessel ".' 


216  HIGHER  ALGEBRA 

17.  A  regiment  received  orders  to  send  216  men  on  garrison 
duty,  each  company  sending  the  same  number  of  men ;  but  before 
the  detachments  marched,  three  entire  companies  were  sent  on 
other  service,  and  it  was  then  found  that  each  remaining  company 
would  have  to  send  12  men  additional  to  furnish  the  required  216. 
How  many  companies  were  in  the  regiment,  and  how  many  men 
did  each  remaining  company  send  on  garrison  duty  ? 

18.  Two  trains  simultaneously  leave  A  and  B,  which  are  81 
miles  apart,  and  pass  each  other  in  1  hour.  The  train  from  A 
reaches  B  27  minutes  earlier  than  the  one  from  B  reaches  A. 
Find  the  time  of  each  train. 

19.  A  boat  goes  along  a  straight  reach  of  a  canal  at  6  miles  an 
hour.  A  person  living  4  miles  from  the  canal  sets  out,  three 
quarters  of  an  hour  before  it  is  due  at  it  nearest  point  to  his  resi- 
dence, to  catch  the  boat.  If  he  goes  4  miles  an  hour,  find  how  far 
below  the  nearest  point  of  the  canal  is  the  point  toward  which  he 
must  direct  his  course,  in  order  that  he  may  reach  it  just  with  the 
boat. 

20.  A  starts  at  10  a.m.  to  walk  from  P  to  Q,  and  B  starts  at 
10 :  24  a.m.  to  walk  from  Q  to  P.  They  meet  6  miles  from  Q.  B 
stops  1  hour  at  P,  and  A  stops  2  hours  and  54  minutes  at  Q,  and 
returning  they  meet  midway  between  P  and  Q  at  6 :  54  p.m. 
Find  the  distance  from  P  to  Q. 

348.  Theorem.  If  all  the  terms  of  an  equation  that  is  integral 
with  reference  to  the  unknown  quantity  be  transposed  to  the  first 
member,  and  the  resulting  polynomial  be  resolved  into  factors  con- 
taining the  unknown  quantity,  the  values  obtained  by  placing  these 
factors  in  turn  equal  to  0  are  the  roots  of  the.  original  equation. 

Dem.  By  hypothesis  the  second  member  of  the  equation  is  0. 
Now  since  any  finite  quantity  multiplied  by  0  is  0,  the  placing  of 
any  one  of  the  factors  equal  to  0  will  render  the  first  member 
0,  and  the  equation  will  be  satisfied.  Hence  the  values  which 
render  the  factor  0  will  be  roots  of  the  original  equation. 

If  the  factors  are  of  the  first  degree  in  the  form  x  —  a,  x—  b, 
x  —  c,  •••  x  —  I,  the  equation  is 

(x  —  a)  (x  —  6)  (x  —  c)  •  •  •  (x  —  I)  =  0, 


QUADRATIC  EQUATIONS  217 

and  from  x  —  a  =  0,  x  —  b  =  0,  x  —  c  =  0,  etc.,  we  have  by  trans- 
position x  =  a,  x  =  b,  x  =  c,  etc. 

349.  Cor.  Conversely,  to  form  an  equation  having  given  roots, 
it  is  but  necessary  to  subtract  the  roots  from  x  and  place  the  product 
of  the  remainders  equal  to  0. 

Thus,  to  form  an  equation  whose  roots  are  1,  8,  and  —  5,  i.e.,  an  equation 
which  is  satisfied  when  x  =  1,  x  =  3,  and  x  =  —  5,  we  have  by  transposition 
£  —  1=0,  x  —  3  =  0,  and  x  +  5  =  0,  and  by  multiplying  these  equations 
together  (using  the  method  of  Art.  50),  x3  +  x2  —  17  x  +  15  =  0. 

If  but  two  roots  be  given,  the  required  quadratic  may  be  written  at  once  by 
Art.  346.    Thus,  if  the  roots  are  2  and  3, 

p  =  -  (2  +  3)  =  -  5,  q  =  -  (2  x  3)  =  -6, 

and  the  equation  is  x2  —  5  x  =  —  G. 

EXAMPLES  LXXXVI 
Solve  the  following  by  Art.  348. 

1.  x2  —  5  x  =  —  6. 

Solution.    Transposing,      x2  —  5  x  +  6  =  0. 
Factoring,  (x  -  2)  (x  -  3)  =  0. 

This  is  satisfied  when  x  —  2  =  0,  and  when  x  —  3=0,  giving  x  =2  and  x=3. 

2.  x2-5x  =  U. 

Solution,     x2  -  5  x  -  14  =  (x  +  2)  (x  -  7)  =  0  ;  .-.  x  =  -  2  or  7. 

3.  a2- 6  <b  + 5  =  0.       4.   ar2-  5  £  =  24.  5.   ar*-6x  +  9  =  0. 
6.  ar*  =  49.                     7.  a2  -  7  a;  =  0.  8.  a?  — 12  a?  =  -36. 

Form  the  equations  whose  roots  are  the  following,  either  writ- 
ing the  products  by  Art.  00,  or  the  equations  at  once  by  Art.  346 : 

9.   4  and  -  2.         10.   7  and  3.  11.    —  8  and  5. 

12.    -6  and -9.     13.    y/S and—  VS.     14.   2+V5and2-V5. 

15.  By  a  misprint  in  an  examination  paper  the  absolute  term 
of  a  quadratic  equation  in  the  form  x2  +px  =  <y  is  made  -f-  instead 
of  — ,  thus  making  the  roots  12  and  —  2.  What  are  the  roots  of 
the  equation  in  the  copy  ? 


218  HIGHER   ALGEBRA 

16.  Two  boys  attempt  to  solve  a  quadratic  equation.  After 
reducing  it  to  the  form  x2  +  px  =  q,  one  of  them  has  a  mistake 
only  in  the  absolute  term,  and  finds  the  roots  to  be  1  and  7 ;  the 
other  has  a  mistake  only  in  the  coefficient  of  x,  and  finds  the  roots 
to  be  —  1  and  —  12.     Find  the  roots  of  the  correct  equation. 

17.  Find  the  relation  that  must  exist  between  p  and  q  in  order 
that  one  root  of  x2  -f  px  -f  q  —  0  may  be  double  of  the  other. 

18.  What,  condition  will  make  one  root  of  x2  -{-  px  =  q  the 
reciprocal  of  the  other? 

350.  Art.  348  has  not  been  introduced  in  connection  with 
affected  quadratic  equations  for  the  purpose  of  replacing  the 
method  of  solution  given  in  Art.  344,  but  for  the  purpose  of 
showing  how  roots  are  sometimes  gained  and  sometimes  lost  in 
transforming  equations,  and  for  providing  against  such  gain  or 
loss. 

351.  Theorem.  Multiplying  an  equation  by  a  factor  containing 
the  unknown  quantity  introduces  into  the  resulting  equation  the  root 
or  roots  obtained  by  placing  the  factor  equal  to  0 ;  and,  conversely, 
removing  from  an  equation  a  factor  containing  the  unknown  quantity 
removes  from  the  equation  the  root  or  roots  obtained  by  placing  the 
factor  equal  to  0. 

Dem.  1st.  The  terms  all  being  transposed  to  the  first  mem- 
ber, the  second  member  is  0;  and  if  the  introduced  factor  be 
made  equal  to  0,  the  product  will  be  0,  and  the  new  equation 
will  be  satisfied.  Hence  the  values  which  render  the  introduced 
factor  0  are  roots  of  the  new  equation. 

2d.  If  an  equation  be  divided  by  a  factor  containing  the 
unknown  quantity,  it  loses  those  values  of  the  unknown  quantity 
which,  by  rendering  that  factor  0,  satisfy  the  original  equation. 

352.  Sch.  In  clearing  of  fractions  an  equation  having  the 
unknown  quantity  in  the  denominator  of  some  of  the  terms,  to 
avoid  the  introduction  of  roots  that  do  not  belong  to  the  original 
equation,  sometimes  called  Extraneous  Roots,  care  should  be  taken 
to  multiply  by  the  lowest  common  denominator;   and  when  the 


QUADRATIC  EQUATIONS  219 

degree  of  an  equation  is  reduced  by  dividing  out  a  factor,  the 
roots  obtained  by  placing  this  factor  equal  to  0  must  be  retained 
as  roots  of  the  original  equation. 

ILLUSTRATIVE    EXAMPLES 

1.  Let  it  be  required  to  find  the  roots  of 

5  2     =     1 

x2  —  1      x  +  1      it'  —  1 

When  we  clear  of  fractions  by  multiplying  by  the  1.  c.  m.  of  the  denomi- 
nators and  reduce,  we  have 

3(x-2)=0, 

and  x  —  2  is  the  only  factor  that  can  be  placed  equal  to  0,  giving  x  =  2  as  the 
only  root. 

When  we  clear  of  fractions  by  multiplying  by  the  product  of  all  the 
denominators  and  reduce,  we  have 

3(x2-l)(x-2)=0, 

which  is  satisfied  not  only  for  x  —  2  =  0,  but  also  for  x2  —  1  =  0,  giving  x  —  2, 
1,  or  —  1.  The  roots  1  and  —  1  do  not  belong  to  the  original  equation,  but 
entered  by  the  introduction  of  the  extra  factor  x2  —  1  in  clearing  of  fractions. 

2.  Let  it  be  required  to  find  the  roots  of 


a2  _  G  k  -  7  =  3  V  ar*  -  6  a  -  7 . 


Dividing  by  vx2  —  6  x  —  7,  we  have 


Vx2  -  6  x  -  7  =  3. 

Squaring  both  members  and  solving,  we  have  x  =  8  or  —  2.  These  num- 
bers satisfy  the  original  equation,  but  so  also  do  7  and  —  1.  How,  then,  did 
these  two  roots  escape  ? 

The  original  equation  may  be  put  in  the  form* 


Vx2  -  6  x  -  7(  Vx2  -  Ox  -  7  -  3)=  0, 

and  this  equation  is  satisfied  by  placing  either  factor  equal  to  0,  and  each  of 
these  equations  gives  two  roots.  Two  of  them,  7  and  —  1,  were  lost  in  divid- 
ing out  the  factor  Vx2  —  6x  —  7. 

If  we  choose,  we  may  free  the  equation  of  radicals,  giving  an  equation  of 
the  4th  degree,  viz., 

^4  _  12  x3  +  13  x2  +  138  x  +  112  =  0, 
and  factor  by  Art.  101,  giving 

(a;_8)(x  +  2)(x-7)(x  +  l)=0, 
the  four  roots  of  which  are  seen  to  be  8,  —  2,  7,  and  —  1. 


220  HIGHER  ALGEBBA 

353.  In  verifying  a  radical  equation  the  double  sign  of  a  radi- 
cal term  is  not  always  admissible.  In  the  last  illustrative  ex- 
ample, if  8  be  substituted  for  x  in  the  equation 


a*  ^  6  x  _  7  s»  3  Vaj2  -6a>-7, 
the  first  member  becomes  9  and  the  second  member  becomes 

3V9  =  3(±3)  =  ±9; 

and,  as  -j-  9  cannot  equal  —  9,  only  the  upper  sign  is  admissible. 
The  same  is  true  when  —  2  is  substituted ;  but,  when  7  or  —  1  is 
substituted,  both  signs  are  admissible,  as  in  each  case  each  mem- 
ber vanishes.  In  such  an  expression  as  x  —  2  =  V#  +  H  there 
are  really  two  equations,  viz., 


a-2  =  +  Va  +  ll  and  x  —  2  =  —  Vs-j- H. 

Each  gives,  when  squared  and  reduced,  x2  -f-  x  =  12 ;  whence, 
x  =  3  or  —  4.  One  of  these  values  satisfies  the  first  of  the  given 
equations,  the  other  satisfies  the  second,  and  both  satisfy  the 
equation  that  results  from  squaring. 

354.  When,  to  free  it  of  radicals  or  for  other  purpose,  an  equa- 
tion is  squared,  there  is  no  way  of  indicating  in  the  resulting 
equation  itself  that  there  is  any  restriction  in  the  matter  of 
signs.  For  example,  if  we  square  both  members  of  x  =  3,  giving 
x2  =  9,  and  then  solve  this  equation,  we  obtain  x  =  3  or  —  3. 
If,  however,  we  transpose  before  squaring,  we  have  x  —  3  =  0, 
x2  —  6x  +  9  =  0;  whence,  x2  —  6  x  =  —  9  and  x  =  3  ±  0  =  3. 
Hence,  additional  roots  may  or  may  not  be  introduced  into  an 
equation  by  squaring  it.  It  follows  that  tvhen  an  equation  is 
rendered  rational  by  squaring  it,  some  of  the  roots  may  have  to  be 
rejected.  A  trial  of  the  roots  in  the  original  equation  will  de- 
termine what  ones  are  to  he  retained. 

EXAMPLES   LXXXVII 

Solve  the  following,  rejecting  extraneous  roots : 
1.   V5  x  +  1  =  V#  +  1  +  2. 


2.    V7  x  +  14  =  V2  x  +  6  +  V#  -f  4. 


QUADRATIC   EQUATIONS  221 


3.   2Vx2-Vx  +  18  -  Var  -  4  a;  -  12 


4.    V2  a:2  +  7x-  d-Vx2  -  5  a;  +  4  =  Vx2  -  1. 


5.    V2  a,-2  +  10  x  +  8  -  Var>  +  6  x  +  5  =  Va?  +  1. 


6.   2Va;  +  V4aj+V7aj  +  2  =  l. 


7.    Vaa;  +  &*  +  Vbx  +  a2  =  a  —  6. 


8.    V(a?-l)(aj-2)+V(a;-3)(a;-4)=V2. 


9.   2Va^  +  -4:  =  5.  10.   *-Vs  +  l  =  _5, 

Va?  a;  +  VaT+T     H 

1  +  6  Va?       Va? 

12.  — 1 + I -& 

V2  4-a?-V2      V2^^"  +  V2       x 


13.   ^±^^  =  (x-2)2.       14.       Vl 


Va^-9  1+vT+a"     1-VT^ 

15.    V4-hV'2^?^  =  ^±i.     16.   a?+Va^I4_a;-V^l=98t 
2  x—^/x2—!     x4-Var*—  1 


CHAPTER   XVI 

SOME  HIGHER  EQUATIONS 

PURE   EQUATIONS 

355.  A  Pure  Equation  is  an  equation  in  which  the  unknown 
quantity  is  affected  with  but  one  exponent. 

356.  Prob.     To  solve  any  pure  equation. 

Rule.  Reduce  to  the  form  in  which  the  affected  unknown  quan- 
tity with  coefficient  unity  shall  constitute  one  member  of  the  equation  ; 
then  perform  upon  both  members  the  operations  necessary  to  make 
the  exponent  of  the  unknown  quantity  unity. 

Dem.     1st.    When  the  form  after  reduction  is  xm  =  a. 

By  extracting  the  mth  root  of  both  members  we  have  x  —  Va. 

m 

2d.    When  the  reduced  form  is  xn  =  a. 

By  extracting  the  mth  root  of  both  members  and  raising  to  the 
?ith  power  both  members  of  the  resulting  equation,  we  have 

x  =  ( Va)n. 

3d.    When  the  reduced  form  is  x  m  =  a. 

By  Art.  147,  2d,  this  is  the  same  as  —  =  a:  whence  xm  =  -,  and 

'      '  xm        '  a 


=  \!a 


357.  Note.  Wlien  both  evolution  and  involution  are  neces- 
sary, as  in  freeing  of  a  fractional  exponent,  it  is  expedient  to  use 
evolution  first,  thus  avoiding  large  numbers. 

It  will  be  shown  in  a  subsequent  part  of  the  work  that  an 
equation  has  as  many  roots  as  indicated  by  its  degree.  In  the 
following  examples  only  those  roots  that  are  obtained  by  the 
above  process  are  required. 

222 


EQUATIONS  IN   THE  QUADRATIC  FORM  223 


3.  a*  =  216. 
6.  a*  =  243. 
9.    </**•  =  256. 


EXAMPLES  LXXXVIII 

Solve  the  following 

1.   a4  =  256. 

2.    0>=s243; 

4.    fflaSl. 

5.   a*=625. 

7       ™-3  —     2  7 

o      T~i  —    8  2,. 

358.  Tlie  ride  of  Art.  356  applies  when,  instead  of  a  simple 
quantity,  as  x,  there  is  a  group  of  terms  affected  as  a  whole  with  an 
exponent. 

EXAMPLES  LXXXIX 
Solve  the  following : 

1.  a6-6a4  +  12a2-8  =  0. 

Solution.  This  is  not  a  pure  equation  with  reference  to  x,  but  is  a  pure 
equation  with  reference  to  X1  —  2.  Kegarding  x2  —  2  as  the  unknown  quan- 
tity, we  have 

x6  -  6  x*  +  12  x2  -  8  =  (x2  -  2)3  =  0, 

whence  x2  —  2  =  0,  and  x  =  ±  V2. 

2.  (o3-6)5  =  32.  3.    (x2  -  Gx  -  2)3  =  125. 
4.    (a? +  2)*  =  81.  5.    (x-2  +  i)"3  =  8. 

6.  Xs  +  9x2  +  27z  +  27  =  0.        7.    a6 -12 a?4 +  48 a2  =  64. 

8.  a,*4-10ar}  +  35x2-50a  +  25  =  0. 

9.  64  a6  +  96^  -  96  x4  -  136  a3  +  72  or*  +  54  a  -  27  =  0. 
10.  (a  +  3)6=(2a2  +  10a  +  4)3. 

EQUATIONS   IN  THE   QUADRATIC   FORM 

359.  An  equation  is  in  the  Quadratic  Form  when  the  unknown 
quantity  has  but  two  exponents,  one  of  which  is  twice  the  other. 

360.  Prob.  To  solve  an  equation  in  or  reducible  to  the  quadratic 
form. 

Solution.  The  reduced  general  form  being  x2m+pxm  =  q,  we 
may  write  it  (xm)2  +  pxm  =  q,  which  is  an  affected  quadratic  equa- 


224  HIGHER  ALGEBRA 

tion  with  reference  to  xm.    Regarding  xm  as  the  unknown  quantity, 
we  have,  by  Art.  344, 

*m  =  -T>±\]j  +  q, 


mM 


whence  x  =  \  — ^  ±  \l~  +  q. 

EXAMPLES   XC 

Solve  the  following : 

1.  x6  +  7ar>  =  8. 

Solution.     Writing  the  equation  in  the  form 

(x3)a  +  7  x3  =  8, 

it  is  seen  to  be  an  affected  quadratic,  not  with  reference  to  x,  but  with  refer- 
ence to  x3.  Regarding  a3  as  the  unknown  quantity  and  solving  by  Art.  344, 
we  have 

x3=-f  ±f  =  1  or  -8, 

whence  x  —  1  or  —  2. 

2.   x4-5x2  =  -4:.  3.   x*-6x2  =  -5.        4.   x4-8x*  =  9. 

5.   a4-2oV  =  -144.         6.   xG-{-6xi  =  l(j.  7.   x?-x%  =  56. 

8.   x*  +  5  J  =  22.  9.   a^  +  a;*  =  756.       10.  x*  +  —  =  —  • 

2  a*      4 

11.   ox"4  -11  x2  =  -6.  12.   a#  +  sr*  =  1056. 

13.   A-  +  A  = 18-  14.  3  *"#?  +  ^=16. 

15.   x*  —  x~*  =  f .  16.   x-s  +  x?  =  *f.. 

361.  The  solution  of  Art.  360  applies  when,  instead  of  a  simple 
quantity,  as  x,  there  is  a  group)  of  terms  having  as  a  whole  only  two 
exponents,  one  of  which  is  twice  the  other. 

EXAMPLES  XCI 
Solve  the  following: 


1.   arJ_4a-6Vx2-4x-5  =  -3. 
Solution.     Subtracting  5  from  both  members,  we  have 
x2  -  4  x  -  5  -  6vV2-  4x-  5  =  -  8. 


EQUATIONS  IN  THE  QUADRATIC  FORM  225 

Here  we  have  a  group  of  terms,  minus  6  times  the  square  root  of  the 
group,  or,  what  is  the  same  thing,  a  group  of  terms  with  the  exponent  1, 
minus  6  times  the  same  group  with  the  exponent  \.  Hence,  regarding 
Vx'2  —  4  x  —  5  as  the  unknown  quantity  and  solving  as  an  affected  quadratic 
(Art.  344),  we  have 

Vx2  -  4  x  -  5  =  3  ±  1  =  4  or  2. 

Squaring,  etc.,  x'2  —  4  x  —  5  =  1C  or  4, 

x2  -  4  x  =  21  or  9, 

x  =  2±5or2±  Vl3 ; 

.-.  x  as  7  or  -  3  or  2  ±  \/l3. 

The  last  two  roots  do  not  satisfy  the  original  equation. 

2.  2x*  -  5x  -2V2x>-  5a  =  15. 


3.   ^-2a-|-6Va2-2a;  +  5  =  ll. 
4.  3 (a +  7)* +  3 (a +  ?)"*  =  10.  5.   #-3VaT+T6  =  -6. 


6.   ajl-V&*-9  =  21. 


7.   2ar2-5z  +  5V2arJ-5a;  +  6  =  33. 


8.  a2  +  3z-V2arJ  +  6a;  +  l  =  l. 

9.  Vx  +  16  -f -</«  +  16  =  6. 

10.   a2+Va2-7a;  +  8  =  7a  +  4. 


11.   2«2-7a;  +  2V2^-7a;+.6=-6. 


12.   2arJ  +  6V2arJ-3a  +  2  =  3z  +  14. 


13.   2a>2-4;c  +  3Vari-2z  +  6  =  15. 

ar*  'a? 


14.   fx-^j+7fx--\  =  12^.  15.   ^  +  1  +  3:  +  - 


16    ^-Sx-S  1  =6 

7  ^-Saj-a     i 


17.   aj*/fl+JLY_(3{Bl  +  »)  =  70. 

J^+Jl-i=:Z.  19-    l+Jl-^=Jl+*. 

*        x      *        x  \         a;      \        a 


ia 

downey's  alo. — 15 


226  HIGHER  ALGEBRA 

EQUATIONS    WITH   INTEGRAL   ROOTS 

362.  An  equation  containing  one  unknown  quantity  is  said  to 
be  in  the  Normal  or  Typical  Form  when  the  exponents  are  all 
positive  integers,  the  coefficient  of  the  highest  power  is  1,  and 
the  other  coefficients  are  integers. 

This  form  is  called  normal  or  typical  because,  as  shown  in  a 
subsequent  part  of  the  work,  every  equation  having  rational  coef- 
ficients can  be  reduced  to  it. 

363.  Prob.  To  find  the  integral  roots  of  an  equation  in  the 
typical  form. 

Solution.  If  the  polynomial  resulting  from  transposing  all 
the  terms  to  the  first  member  has  integral  factors  of  the  first 
degree,  they  may  be  found  by  the  process  of  Art.  101.  Now,  by 
Art.  348,  the  roots  obtained  by  placing  these  factors  equal  to  0 
are  roots  of  the  original  equation. 

If  all  the  roots  but  two  are  found  by  this  process,  the  remain- 
ing two,  even  if  surd  or  imaginary,  may  be  found  by  placing  the 
remaining  factor,  which  is  quadratic,  equal  to  0. 

364.  Cor.     Some  or  all  of  the  roots  may  be  equal. 

Note.  In  solving  the  following  examples  carefully  observe 
the  instructions  of  Art.  102. 

EXAMPLES   XCII 

Find  the  roots  of  the  following : 

1.   x>-xA-  13a?  +  13a?2  +  36a;-  36  =  0. 

Operation.  Proceeding  as  in  Art.  101,  using  the  smaller  divisors  of  30 
(only  factors  of  the  absolute  term  need  be  tried  in  any  case),  we  have  the 
following : 

xh  _  x\  _  13  xs  +  13  x-2  +  36  x  -  36  I      1 


0- 

-13 

0 

36 

0  | 2 

2- 

-    9 

-18 

0 

Li 

5 

6 

0 

LzJ 

S 

0 

|-3 

0 

EQUATIONS    WITH  INTEGRAL   ROOTS 


227 


5.   ar*  +  2x2-23#  -60  =  0. 


Therefore  the  roots  are  1,  2,  3,  —  2,  —  3.  This  is  because  the  factors  are 
x  —  1,  x  —  2,  x  —  3,  x  +  2,  and  x  +  3,  and  when  these  are  placed  equal  to  0 
(Art.  348),  we  have  by  transposition 

x  =  1,  x  =  2,  x  =  3,  x  =  -  2,  x  =  -  3. 
2.   ic3-  9ar>  +  26z  -24  =  0. 
4.   ^-8x2-fl3x-6  =  0. 

6.   ar3-a;2-8.T  +  12  =  0.  7.   ar5  -  5x2  -  8a;  +  48  =  0. 

8.   ar3  +  8arJ  +  20a;  +  l6  =  0.  9.   ar3-13ar  + 47a; -35  =  0. 

10.  a;4  -3ar3-  14a;2  +  48a; -32  =  0. 

11.  rf-lla?  +  l&x-8  =  0. 

Sug.     Supply  the  missing  term,  with  coefficient  0. 

12.  a;4-45arJ-40o;  +  84  =  0. 

13.  a?4 +  13  a? +  33  or8 +  31  a; +  10  =  0. 
Query.     Why  is  it  unnecessary  to  try  any  positive  numbers  ? 

14.  a*-3ar5  +  6ar}-3arJ-3a;  +  2  =  0. 


Operation 

x«-3x£ 

+  0  x4  +  6  x3  - 

-3x2 

-3x 

+  21      1 

-2 

-2 

4 

1 

-2 

01     i 

-1 

-3 

1 

2 

0 

l_l 

0 

-3 

-2 

0 

[=1 

-1 

-2 

0 

bi 

-2 

0 

L^ 

0 

Hence  the  roots  are  1 , 

1,1, 

-1,-1 

,2. 

15.  a?-2a;4-15ar3  +  8a;2  +  68a;  +  48  =  0. 

16.  x"  -  13  a;4  +  67  a?  -  171  x2  +  216  a;  -  108  =  0. 

17.  a^5  +  3ar5-6a;4-6ar3  +  9arJ  +  3a;-4  =  0. 

18.  a;7  +  5a^  +  6ar5-6a;4-15ar3-3arJ  +  8a;  +  4  =  0. 

19.  ari-8a;4  +  21ar3-16a,-2-10a;  +  12  =  0. 


228  HIGHER  ALGEBRA 

Operation 

X5  _  8  x±  +  21  x*  -  16  x2  -  10  x  +  12  (_J_ 
-7  14        -2-12  0(_2 

-5460  L§ 

-2        -2  0 

Further  trial  by  this  process  fails  to  give  the  other  two  roots.  (There 
must  be  five  roots  (Art.  357),  since  the  equation  is  of  the  fifth  degree.) 
However,  the  factors  of  the  polynomial  are  x  —  1,  x  —  2,  x  —  3,  and 
x2  —  2  x  —  2,  the  last  row  of  numbers,  1,  —  2,  —  2,  furnishing  the  coefficients 
for  the  last  one  (Art.  81).    Placing  the  quadratic  factor  equal  to  0  (Art.  348), 

we  have 

x2  -  2  x  -  2  =  0, 

or  x2-2x  =  2, 

whence  x  =  I  ±  VS. 

Hence  the  roots  are  1,  2,  3,  1  ±  V3. 

20.    ^-6^  +  10^-8=0.  21.    a3 -3  a2 +  £  +  2  =  0. 

22.   aj*  -6  ar*  +  24 x  -16  =  0.  23.  a4  -  4  a?  -  8x  +  32  =  0. 

24.  ^-9^  +  17^  +  27^-60  =  0. 

25.  ar5-3a4-9ar5  +  21a?2-10:r  +  24  =  0. 

26.  .^5-4x4-16ic3  +  112^2-208«  +  128  =  0.    . 

27.  x6-7a,-5  +  lla;4-7^  +  14a2-28a;  +  40=:0. 

365.  The  method  of  Art.  363  applies  also  when  the  coefficient 
of  the  first  term  is  not  1 ;  but  in  that  case,  if  there  is  no  mono- 
mial factor  common  to  all  the  terms,  the  rational  roots  are  not  all 
integral  divisors  of  the  absolute  term.  We  should  find  first  any 
integral  roots  the  equation  may  contain,  and  find  the  remaining 
root  or  roots  by  placing  the  remaining  factor  equal  to  0. 

EXAMPLES  XCIII 
Find  the  roots  of  the  following : 
1.   3x4-8^-llx2  +  28a?-12  =  0. 

Operation 

3  X4  _  8  x3  -  11  x2  +  28  x  -  12  J 1 

-  5      -  16  12  0  | 3 

4-4  0  1  -2 

-2  0 


EQUATIONS    WITH  INTEGRAL  ROOTS  229 

The  last  factor  of  the  polynomial  is  seen  to  be  3  x  —  2  ;  and  by  Art.  348, 
3  x  -  2  =  0, 
whence  x  =  §. 

Hence  the  roots  are  1,  3,  —  2,  f . 
2.   6^-17  z4-25ar3  +  55x2  +  39a-18  =  0. 

Operation 

G  x5  -  17  x4  -  25  x3  +  55  x2  +  39  x  -  18  | 2 

_    5      _  35      _  15  9  0  | 3 

13  4-3  0  |  -1 

7-3  0 

By  Art.  348,  6  x2  +  7  x  -  3  =  0, 

whence  x2  +  £  x  =  \, 

and  x  =  -  T73  ±  H  =  i  <*  -  !• 

Hence  the  roots  are  2,  3,  —  1,  |,  —  f . 

3.  5^-22^ +  15^  +  18  =  0. 

4.  6arJ-29.T2-6a>+5  =  0. 

5.  6z4  +  5arl-25arJ-10a;  +  24=:0. 

6.  4af-28z4  +  57arJ-8x2-67.T  +  30  =  0. 

366.  By  Art.  349,  produce  the  equations  whose  roots  are  the 
following,  performing  the  multiplications,  as  far  as  practicable, 
as  in  Art.  59 : 

1.   1,   -  3,  4.  2.   1,  3,   -  2,   -  4. 

3.   V2,  -V2,  -1,  3.  4.   1,  2,  2,  -3,  4. 

5.   2,  3,  4,  -  1,  -  5.  6.   1,  3,   -  2,   -  2,  -  2. 

7.    -3,  2  +  V^l,  2-V==~l.       a    ±V:r2,   ±V5. 
9.   1  ±V^2,  2±V:::3.  10.  f,  2,  V3,  -VS. 

Solution,     (x  -  § )  (x  -  2)  (x  -  V3)  (x  +  V3)  =  0, 

(2x-3)(x-2)(x2-3)  =  0,  and  2  x*  -  7  x3  4-  21  x  -  18  =  0. 

2  x  —  3 
Query.     Why  may  the  denominator  of  — —  —  be  dropped  ? 

U9     4.1     3  12     1      _I     _1     _3 


CHAPTER   XVII 

SIMULTANEOUS   EQUATIONS   OF  THE   SECOND   DEGREE 
WITH  TWO  UNKNOWN  QUANTITIES 

367.  A  Homogeneous  Equation  is  an  equation  in  which  all  the 
terms  are  of  the  same  degree.* 

Thus;  8  x2  -  2  xy  =  4  y2,  x2  -  4  xy  +  5  y2  =  0,  ax*  +  bx2y  +  cxy2  +  dys  =  0, 
are  homogeneous  equations. 

368.  A  Symmetrical  Equation  is  an  equation  in  which  the 
unknown  quantities  may  change  places  without  affecting  the 
equality. 

Thus, 
Sx2  +  3y'2  -  x  -  y  =  5,  2  x2  +  2 y2  -  3 xy  =  7,  ax2  +  ay2  +  &x?/  +  cz  +  ty  =d, 
are  symmetrical  equations. 

369.  Theorem.  The  solution  of  two  equations  of  the  second  degree 
with  two  unknown  quantities  requires,  in  general,  the  solution  of  a 
biquadratic. 

Dem.  Two  general  equations  of  the  second  degree  with  two 
unknown  quantities  (such  equations  must  provide  for  all  terms 
that  can  possibly  occur)  have  the  forms 

ax2  +  bxy  +  af  +  dx  +  ey  +  f  =0,  (1) 

a'x2  +  b'xy  +  c'y2  +  d'x  +  e'y  +/'  =  0.  (2) 


From(l)       ,,  -»»±*±  JQ*^^&±a±L 

2  a         ^      4a2  a 

The  substitution  of  this  in  (2)  will  give  terms  in  y2  and  a  radical  of 
the  second  degree.  Then  the  rationalization  of  this  equation  will 
require  the  squaring  of  a  polynomial  containing  y2,  and  the  result- 
ing equation  will  be  of  the  fourth  degree. 

*  Some  writers  apply  the  name  to  equations  in  which  all  the  terms  except 
an  absolute  term  are  of  the  same  degree. 

230 


SIMULTANEOUS  EQUATIONS  231 

370.  Although  the  elimination  of  one  of  the  two  unknown 
quantities  from  two  equations  of  the  second  degree  results,  in 
general,  as  shown  above,  in  a  biquadratic,  there  are  many  cases  in 
which,  owing  to  the  absence  of  some  of  the  terms,  the  solution 
may  be  effected  by  the  use  of  a  quadratic,  or  by  the  use  of  two 
simple  equations  with  two  unknown  quantities.  The  most  useful 
of  the  various  methods  are  here  given. 

Case  I 

371.  When  one  of  the  equations  is  of  the  first  degree. 

Rule.  Find  from  the  simple  equation  the  value  of  one.  of  the 
unknown  quantities  in  terms  of  the  other  and  known  quantities. 
Substitute  this  in  the  other  equation  and  solve  in  the  usual  way  the 
resulting  quadratic. 

Dem.  The  only  feature  of  this  rule  needing  proof  is  that  the 
equation  resulting  from  eliminating  one  of  the  unknown  quanti- 
ties is  a  quadratic. 

The  general  form  of  an  equation  of  the  second  degree  with  two 
unknown  quantities  is 

ax1  +  bxy  +  cy-  +  dx  +  ey  +/=  0. 

The  general  form  of  an  equation  of  the  first  degree,  with  two 
unknown  quantities,  is 

mx  +  ny  +  p  =  0. 

x  From  the  latter,  x  =  ~mJ~V, 

m 

which,  substituted  in  the  former,  gives  no  term   containing   a 
higher  power  of  y  than  the  second. 

EXAMPLES   XCIV 
Solve  the  following : 

7  x2  -  8  xy  =  159,  f  x--  2  xy-y2  =  l, 

ox  +  2y  =  l.  '   \x  +  y  =  2. 

f*  +  ?'  =  4'  1^+4^  +  23  +  37=17, 

-  +  -  =  1-  '    [2x-y  =  0. 

x      y 


232  HIGHER   ALGEBRA 

a  .  b 


x      y      10 


x      y 

a2      b2      n 


Although  the  last  two  examples  contain  no  equation  of  the  first  degree 
with  reference  to  x  and  y,  the  first  equation  of  each  is  of  the  first  degree 
with  reference  to  the  reciprocals  of  x  and  y,  and  the  second  equation  of  each 
contains  the  squares  of  these  reciprocals.     We  may,  therefore,  regard  -  [or 

in  the  0th  -  )  as  the  unknown  quantity,  and  solve  accordingly. 
x) 

Case  II 

372.  When  each  equation  has  only  one  term  of  the  second  degree 
and  these  terms  are  similar. 

Rule.  Eliminate  the  terms  of  the  second  degree,  and  combine 
the  resulting  equation  of  the  first  degree  with  either  of  the  original 
equations,  as  in  Case  I. 

EXAMPLES  XCV 

Solve  the  following : 

t  f  _j_  3  x  _  4  y  =  6,  |  6  x2  +  9  x  -  2  y  =  15, 

(2xy-2x  +  ±y  =  ±,  (  3xy  +  6x  -  3y  =  10, 

{  xy  -+-  x  +  6  y  =  6.  1  6  xy  +  2  #  —  y  =  10. 

Case  III 

373.  WVie/i  o?ie  o/  tfie  equations  is  homogeneous. 

Rule.  Find  from  the  homogeneous  equation  the  value  of  one 
unknown  quantity  in  terms  of  the  other  and  knoivn  quantities. 
Substitute  this  value  in  the  other  equation  and  solve  the  resulting 
quadratic. 

Dem.     The  general  form  of  a  homogeneous  equation  of  the  sec- 
ond degree  with  two  unknown  quantities  being 
ax2  +  bxy  +  cy2  =  0, 
by  ,    y 


we  have  x  =  —  7^-±  ~Vb2  —  4  ac, 

2a     2a 

or  x  =  ~(—  b±-Vb2  —  4ac). 

2ax  J 


SIMULTANEOUS   EQUATIONS  233 

As  this  is  of  the  first  degree  with  reference  to  y,  no  term  higher 
than  the  second  degree  can  result  from  substituting  this  value  of 
x  in  an  equation  of  the  second  degree. 

EXAMPLES   XCVI 
Solve  the  following : 

x2  +  y2  +  2x  =  12,  |  x2  +  5xy  +  6y2  =  180, 

3ar>  +  2xy-y2  =  0.  *    [  x2  +  xy  -  6y2  =  0. 


f2s"-2a;  +  y  =  13,  4 


2x2-2x  +  y  =  13,  •     ($x2-6xy  +  y2  =  0, 

4. 

Case  IV 


x2  +  f  +  x  =  6. 


374.    When  both  equations  have  an  absolute  term,  but  are  other- 
wise homogeneous. 

FIRST    METHOD 

Rule.     Eliminate  the  absolute   terms  and  then  proceed  as  in 
Case  III. 

(x2-xy  +  y2  =  21,  (1) 

Example.  \ 

\2xy-y2  =  15.  (2) 

Solution.     Multiplying  (1)  by  5  and  (2)  by  7,  and  subtracting,  we  have 

5x2-    bxy  +    5  j/2  =  105 
14  xy  -    7  y2  =  105 


6  x2  -  19  xy  +  12  y2  =     0 
This  solved  for  x  gives 

x  =  tty±Uy  =  zy°rU'  (3) 

Substituting  the  first  of  these  values  in  (2)  and  solving,  we  have 

y  =  ±  V3. 

Substituting  these  values  of  y  in  that  part  of  (3)  used  in  finding  them,  we 
have 

x  =  3  y  =  ±  8  \/3. 

Substituting  the  other  value  of  x,  viz.,  $  y,  in  (2)  and  solving,  we  have 

y  =±5. 

Substituting  these  values  of  y  in  *A«i  par*  of  (3)  wsed  m  finding  them,  we 
have 

x  =  £  y  =  ±  4. 


Example. 


234  HIGHER  ALGEBRA 

It  should  be  noted  that  the  values  of  x  and  y  occur  in  pairs,  thus  : 

Ja;  =  3V3,  Jx  =  -3V3,  jx  =  4,  fx  =  -4, 

\y  =  y/3.  \y  =  -VS.  U  =  5.  If  =  -  5. 

SECOND    METHOD 

375.  Rule.  Find  from  one  of  the  given  equations  the  value  of 
one  of  the  unknown  quantities  in  terms  of  the  other  and  known 
quantities.  Substitute  this  value  in  the  other  equation  and  solve  in 
the  usual  way  the  resulting  equation,  which  will  always  have  the 
quadratic  form  (Art.  359). 

x2-xy  +  y2  =  21,  (1) 

2xy-f=15.  (2) 

Solution.     From  (2)  x  = 15  +  fc  (3) 

2y 

Substituting  this  value  of  x  in  (1)  and  reducing,  we  have 
f«  -  28  y*  =  -  75, 
whence  y2  =  14  ±  11  =  25  or  3, 

and  y  =  ±  5  or  ±  V3. 

These  values  of  y  substituted  in  (3)  give 

x=±4  or  ±3>/3. 

THIRD    METHOD 

376.  Rule.  Assume  x  =  vy,  and  substitute  in  both  equations. 
By  elimination  form  an  equation  involving  only  v,  and,  solve  for  v. 
Simple  substitution  will  then  determine  x  and  y. 

Dem.  Taking  the  most  general  form  of  the  equations  in  ques- 
tion, we  have 

ax2  -f  bxy  +  cy2  =  m, 

dx2  -f  exy  -\-fy2  =  n. 

Now  if  x  =  vy,  (1) 

where  v  is  simply  the  ratio  of  x  to  y, 

av2y2  +  bvy2  +  cy2  =  m, 

dv2y2  +  evy2  +fy2  =  n ; 

whence  y2  = ™ = 2 (2) 

av2  +  bv  +  c     dvr  +  ev+f 


SIMULTANEOUS  EQUATIONS  235 

As  this  equation  is  of  the  second  degree  with  reference  to  v,  it 
can  always  be  solved. 

When  the  value  of  v  is  substituted  in  either  of  the  equations 
(2),  y  becomes  known ;  and  when  v  and  y  are  substituted  in  (1), 
x  becomes  known. 

Example. 

[2xy  —  y2  =  15. 

Solution.     Assuming  x  =  vy,  (1) 

and  substituting  in  both  equations,  we  have 

21  15 


y       v-2  _  v  +  i      2  v  -  l' 
whence  v  =  3  or  f . 

Substituting  the  first  value  of  v  in  (2), 

2  v  —  1 
whence  y  =  ±  V3. 

Substituting  in  (1),  x  =  vy  =±  Z\/S. 

Substituting  the  second  value  of  v  in  (2), 


2/2  =  ~15T  =  25, 

2  v  —  1 

whence 

y  =  ±5. 

Substituting  in  (1), 

x  =  i>y  =  ±  4. 

Solve  the  following : 

EXAMPLES  XCVII 

far9  +  ^  =  15,  jar9  +  ^+4^  =  6, 

f^  +  ^4-22/2  =  74,  (^+7/  + l=3a*/, 


(2) 


3'  J2a8  +  2xy  +  tf  =  73.  *'  1  2(ojy +  4)  =  3y*. 

jar>  +  ^  +  2/2  =  52,        .  r^-2^-7/^31, 

U-i/-ar  =  8.  *   1  1^  +  2*2/ -2/2  =  101. 

377.  Many  equations  falling  under  Case  IV.,  viz.,  equations 
having  an  absolute  term,  but  otherwise  homogeneous,  admit  of 
shorter  solutions  than  by  any  of  the  three  methods  given  above. 


236  HIGHER  ALGEBRA 

1st.  When  any  multiple  of  one  equation  added  to  or  subtracted 
from  the  other  gives  a  perfect  square. 

Rule.  If  two  simple  equations  are  thus  obtained,  finish  the 
solution  by  any  of  the  methods  of  elimination.  If  but  one  simple 
equation  is  obtained,  finish  the  solution  as  in  Case  I. 

1  xy  =  28.  (2) 

Solution.     Adding  twice  (2)  to  (1),  we  have 
x2  +  2  xy  +  y2  =  121, 
whence  x  +  y  =  ±11.  (3) 

Subtracting  twice  (2)  from  (1),  we  have  • 

x2  -  2  xy  +  y2  =  9, 
whence  x  —  y  =  ±  3.  (4) 

The  combination  of  (3)  and  (4)  gives 

x  =  ±  7  or  ±4, 
y  =  ±  4  or  ±7. 

The  same  process  is  often  applicable  to  equations  above  the 

second  degree. 

(x*  +  3x*y*  =  28,  (1) 

Example.    <    0  0 

lary  +  47/4  =  8.  (2) 

Solution.     The  addition  of  (1)  and  (2)  gives 
xi  +  4  x2y2  +  4y2  =  36, 
whence  x2  +  y2  =  ±  6, 

and  y2  =  ±  6  -  x2.  (3) 

Substituting  this  in  (I)  and  reducing,  we  have 
x^9x2  =  -  14, 
whence  z2  =  ±f±f  =  ±7  or  ±4, 

and  x  =  ±  \/l  or  ±  V-~T,  or  ±  2  or  ±  2V- 1. 

By  substituting  these  values  of  x  in  (3),  y  may  be  found. 

2d.  When  the  first  members  of  the  two  equations  have  a  common 
factor  containing  the  unknown  quantity. 

Rule.  Divide  one  equation  by  the  other,  canceling  the  commoyi 
factor.     Then  proceed  as  in  Case  I. 


SIMULTANEOUS  EQUATIONS  237 

(x2  +  xy  =  35,  (1) 

Example.  ,      ,,  /fV 

(<*.+  /*>  14  (2) 

Solution.     The  equations  may  be  put  in  the  forms 

*(x +  20=35,  (3) 

y(x  +  y)=  14.  (4) 

Dividing  (3)  by  (4),  we  have 

x_35 

2/~14' 

whence  x  =  -— f~  (5) 

14 

Substituting  in  (2)  and  solving, 

V  =  ±2. 

Substituting  these  values  of  y  in  (5), 

*  =  ±  5. 

The  same  process  is  applicable  to  equations  above  the  second 
degree  whose  first  members  have  a  common  factor  containing  the 
unknown  quantity. 

a?  +  3f  =  28,  (1) 


Example. 

*  +  2/  =  4.  (2) 

Solution.    Dividing  (1)  by  (2),  we  have 

x2  -  xy  +  y1  =  7.  (3) 

Proceeding  with  (2)  and  (3)  as  in  Case  I.,  we  find 

x  =  1  or  3, 
y  =  3  or  1. 

EXAMPLES  XCVIII 
Solve  the  following  by  Art.  377  : 

,  x2-?f  =  12,  m    {x2  +  y2  =  n, 


(x2  +  y2  = 

[  xy  =  35. 


xy  +  y2  =  12.  I  xy 

3    ^x2y  +  xy2  =  30,  4    j^  +  ^91, 

'    [x  +  y  =  5.  '  \x  +  y  =  l. 

r  x2-±y2  =  9,  (xi  +  3xy  =  10, 

5'  \xy  +  2y2  =  3.  *    \xy  +  4y2  =  6. 

(xs-tf  =  875,  (xA-\-x2y2-\-y4  =  133J 

7"    [x2  +  xy  +  y2  =  175.  *  1  x2  -  xy  +  y2  =  7. 


238  HIGHER  ALGEBRA 

Case  V 

378.  When  the  equations  are  symmetrical,  and  not  included  in 
any  of  the  preceding  cases. 

Rule.  Assume  x  =  u  +  v  and  y  =  u  —  v,  and  substitute  in  both 
equations.     Then  reduce  and  eliminate  v. 

Dem.  Since  x  and  y  are  by  hypothesis  involved  alike  (Art. 
368),  u  -\-  v  and  u  —  v  will  be  involved  alike.  Hence  for  every 
plus  term  containing  an  odd  power  of  v  there  will  be  an  equal 
negative  term.  Therefore  only  even  powers  of  v  will  remain,  and 
v  can  be  eliminated. 

The  method  is  not  limited  to  equations  of  the  second  degree. 

EXAMPLES  XCIX 
Solve  the  following : 

,txy(x  +  y)  =  30,  (1) 


s 


Xs  +  f  =  35.  (2) 

Solution.     Assume  z  =  u  +  v,  (3) 

and  y  =  u  —  v.  (4) 

Substituting  these  values  in  (1)  and  (2)  and  reducing,  we  have 

us  -  uv2  =  15,  (5) 

and  2  m3  +  6  uv*  =  35.  (6) 

Adding  6  times  (5)  to  (6),  we  have 

8  u*  =  125, 
whence  u  =  f . 

Substituting  this  value  of  u  in  (5)  and  solving,  we  obtain 

Substituting  these  values  of  u  and  v  in  (3)  and  (4) ,  we  obtain 

x  -  |  ±  \  =  3  or  2, 

and  y  —  f  T  i  =  2  or  3. 

In  this  example  a  shorter  process  is  to  add  3  times  the  first  equation  to  the 
second,  extract  the  cube  root  of  the  result,  giving  x  +  y  —  5,  and  substitute 
this  value  in  the  first  equation,  giving  xy  =  6.  Then  the  equations  x  HJ  Jf  as  6 
and  xy  =  6  are  easily  solved. 


SIMULTANEOUS  EQUATIONS 


239 


a*  -f  y*  +  *  +  y  »  20. 

2ar9  +  22/2  =  5^, 
4(a*  +  y)=»y. 


|  a*4  —  9  ar2^2 
1  a;  +  w  =  4. 


9*Y  +  y«  =  l, 


5. 


2/ 

'^2  +  2/2  —  «  —  y  =  14, 

.  a#  4-  x  4-  y  =  14. 


MISCELLANEOUS   EXAMPLES 
EXAMPLES  C 


1. 


Solve  the  following: 

'a*  +  xy  =  12, 
I  xy  4-  2/2  =  2. 

[  a2  -  ?/2  =  45, 
I  a?  —  y  =  3. 


|3ajy-a;-5y  =  8, 
I  a?y  +  a;  —  3  #  =  4. 


5. 


6 


'   \xy  =  12. 


32, 


xy 

x2  +  y-  4-  x  4-  y  =  18, 


11. 


13. 


15. 


17. 


r  a  4-  y  4-  V #  -f  i/  =  6, 
U  +  ^lft 


a-2  +  3a#  =  28, 
a*/  4-  4 1/2  =  8. 

x2-Q>xy  +  y2-2x+l0y  =  -12, 
x2  +  6xy-7y2  =  Q. 

x2-y2  =  60, 
a^  =  16. 

x2  +  xy  =  12, 
it*?/  —  2  2/2  =  1  - 

(x*  +  y*  =  97, 
I  a*  4-  ?/  =  5. 

a?2     y2 
ab 


10. 


12. 


+  -2=io, 
y2 

3. 


x2  4-  xy  =  15, 
xy-f  =  2. 

x*-tf  =  26, 

a*2?/  —  ar?/2  =  6. 

3a-2  +  a-?/-2?/2=162/, 

x2  —  2  2Z2  =  4 1/. 


14 


16. 


18 


■{ 


an/ 

a;  4-  Vxy  +  y  =  19, 
aj»  +  xy  +  /  =  133. 
8 ar3-  27 ^  =  271, 
2a--3y  =  l. 
x2  +  3  a*/  +  2  ?/2  =  63, 

8a#  +  4y2  =  171. 


1^4-3 
13a*2  4- 


240  HIGHER  ALGEBRA 


{tf  +  tf  =  152, 
I  x2  —  xy  +  y2  =  19. 


20. 


|  &  +  y2  +  3xy-  ±(x+y)=-  3, 
I  xy  4-  2  (a?  4*  y)  =  5. 


PROBLEMS   LEADING   TO   SIMULTANEOUS   EQUATIONS   OF  THE 
SECOND   DEGREE   WITH  TWO  UNKNOWN  QUANTITIES 

EXAMPLES  CI 

1.  A  certain  rectangle  contains  300  square  feet;  a  second 
rectangle  is  8  feet  shorter  and  10  feet  broader,  and  also  contains 
300  square  feet.     Find  the  dimensions  of  the  first  rectangle. 

2.  The  area  of  a  rectangular  field  is  300  square  rods,  and  the 
length  of  its  diagonal  is  25  rods.     Find  the  sides. 

3.  A  man  bought  some  horses  for  $  1250.  At  another  time  he 
bought  3  more  than  before,  paying  $  25  less  apiece,  and  they  cost 
him  $  1300.  How  many  horses  did  he  buy  the  first  time,  and  at 
what  price  ? 

4.  The  fore  wheel  of  a  buggy  makes  6  revolutions  more  than 
the  hind  wheel  in  going  120  yards ;  but  the  fore  wheel  of  a  coach, 
each  of  whose  wheels  is  larger  in  circumference  by  1  yard  respec- 
tively, makes  only  4  revolutions  more  than  the  hind  wheel  in  going 
the  same  distance.  What  is  the  circumference  of  each  buggy 
wheel  ? 

5.  In  walking  to  the  summit  of  a  mountain  a  man's  rate  during 
the  second  half  of  the  distance  is  \  mile  per  hour  less  than  during 
the  first  half,  and  he  reaches  the  summit  in  5J  hours.  He  de- 
scends in  3|  hours  at  a  uniform  rate,  which  is  1  mile  per  hour 
more  than  his  rate  during  the  first  half  of  the  ascent.  Find  the 
distance  to  the  summit  and  the  rates  of  walking. 

6.  Two  trains  start  at  the  same  time  from  two  places,  A  and  B, 
168  miles  apart,  and  travel  toward  each  other.  They  pass  in  1  hr. 
52  min.,  and  the  first  reaches  B  \  hour  before  the  second  reaches 
A.     Find  the  speed  of  each  train. 


SIMULTANEOUS  EQUATIONS  241 

7.  A  crew,  rowing  at  half  their  usual  speed,  row  3  miles  down 
stream  and  back  in  2  hr.  40  min.  At  full  speed  they  can  go  over 
the  same  course  in  1  hr.  4  min.  Find  the  rate  of  the  crew  and  of 
the  current. 

a  A  courier,  riding  at  a  uniform  rate,  left  a  station.  Five 
hours  afterward  a  second  followed,  riding  3  miles  an  hour  faster. 
Two  hours  after  the  second  a  third  started  at  the  rate  of  10  miles 
an  hour.  They  all  reached  their  destination  at  the  same  time. 
Find  the  distance ;  also  the  rate  of  the  first. 

9.  A  and  B  are  two  towns  situated  18  miles  apart  on  the  same 
bank  of  a  river.  A  man  goes  from  A  up  to  B  in  4  hours,  rowing 
the  first  half  of  the  distance  and  walking  the  second  half.  In 
returning  he  walks  the  first  half  at  the  same  rate  as  before,  but 
the  stream  being  with  him,  he  rows  1£  miles  per  hour  faster  than 
in  going,  and  covers  the  whole  distance  in  3£  hours.  Find  the 
rates  of  rowing  and  walking. 

10.  A  man  arrives  at  the  railroad  station  nearest  his  home  1\ 
hours  before  the  time  at  which  he  has  ordered  his  carriage  to 
meet  him.  He  sets  out  at  once  to  walk  at  the  rate  of  4  miles  an 
hour,  and  meeting  his  carriage  when  it  has  traveled  8  miles, 
reaches  home  1  hour  earlier  than  he  had  originally  expected. 
How  far  is  his  home  from  the  station,  and  at  what  rate  was  his 
carriage  driven  ? 


i>o\vm:y's  alg.  — 16 


CHAPTER   XVIII 

THEORY  OF  FUNCTIONS 

SECTION  I  — MAXIMA  AND   MINIMA  OF  FUNCTIONS 

379.  An  Arbitrary  Constant  is  a  constant  to  which  any  value 
may  be  assigned,  but  which  maintains  the  same  value  throughout 
the  same  operation  or  discussion. 

An  Absolute  Constant  is  a  constant  which  admits  of  no  change. 

Thus,  in  the  formula  for  the  area  of  a  circle,  wr2,  r  is  an  arbitrary  constant, 
since  it  may  have  any  value  we  choose  to  give  it ;  but  t  ,  always  having  the 
same  value,  viz.,  3.14159  approximately,  is  an  absolute  constant.  When  we 
assign  to  r  any  particular  value,  as  5,  the  radius  becomes  an  absolute  constant. 

Arbitrary  constants  are  represented  by  the  leading  letters  of 
the  alphabet,  and  absolute  constants  by  figures  or  by  letters  which 
always  stand  for  the  same  numbers. 

380.  A  Variable  is  a  quantity  which  may  have  in  the  same 
operation  or  discussion  any  value  within  the  limits  determined  by 
the  conditions. 


Thus,  if  y  =  V25  —  z'2,  and  this  is  the  only  required  relation  between  x  and 
y,  we  may  give  to  x  any  values  whatever  between  —  5  and  +  5  and  find 
corresponding  values  of  y.  Hence  x  and  y  are  variables.  If  x  is  made  less 
than  —  5  or  greater  than  +  5,  y  is  imaginary.  Hence  the  limits  of  x  are  —  5 
and  +  5. 

Variables  are  represented  by  the  final  letters  of  the  alphabet. 

381.  Constants  and  variables  are  not  the  same  as  known  and 
unknown  quantities,  although  the  notation  is  the  same.  In  the 
simultaneous  equations  5  x  -f  2  y  =  7  and  7  x-  —  Sxy  =  159,  x  and 
y  have  two,  and  only  two,  values  each,  and  hence  they  are  constants ; 
but  if  x  and  y  are  required  to  fulfill  only  the  one  condition  ex- 
pressed in  the  first  equation,  5  x  +  2  y  =  7,  we  can  give  to  one  of 
them  any  value  we  please,  and  can  find  for  the  other  such  value 

242 


MAXIMA   AND  MINIMA    OF  FUNCTIONS  243 

as   will   make   the  equation  true.      In   this   case   x  and   y  are 
variables. 

382.  A  Function  of  a  Variable  is  any  expression  which  depends 
upon  that  variable  for  its  value. 

Thus,  y/^b  —  x1  is  a  function  of  x,  inasmuch  as  it  changes  when  x  changes. 
Any  expression  containing  x  is  a  function  of  x.  If  we  have  y  =  V25  —  x'2, 
we  say  y  is  a  function  of  x. 

383.  A  Function  of  Two  or  More  Variables  is  any  expression 
which  depends  upon  those  variables  for  its  value. 

Thus,  interest  on  a  loan  of  money  is  a  function  of  the  principal,  the  rate  of 
interest,  and  the  time  ;  the  volume  of  a  cone  is  a  function  of  the  radius  of  the 
base  and  the  altitude  ;  the  distance  passed  over  by  a  body  moving  from  rest 
with  a  uniformly  accelerated  velocity  (D  =  \ft2)  is  a  function  of  the  acceler- 
ation and  the  time. 

384.  Notation.  A  function  of  x  is  represented  by/(#).  This  is 
employed  not  only  to  represent  any  function  of  a  single  variable, 
but  also  to  represent  a  specified  function  to  avoid  repetition  of  the 
function  itself.  In  this  expression /is  not  a  factor,  but  simply  an 
abbreviation  of  the  word  function. 

When  different  functions  of  the  same  variable  are  brought  into 
the  same  discussion,  the  notation 

/(»)>  /(*),  /"(*)>  /iO)>  M*),  *(*),  *'(*)*  etc., 
is  employed.     These  are  read,  "/function  of  x,"  "/'  function  of 
x,"  "f"  function  of  x"  "f  function  of  aj,"  "/2  function  of  a?," 
"</>  function  of  x,"  "  <f>'  function  of  x,"  etc. 

To  represent  the  same  function  of  different  variables,  the  nota- 
tion f{x),  f{y),  f(z),  /(  -  x),  etc.,  is  employed. 

To  indicate  that  a  constant  has  been  substituted  for  the  variable 
in  a  function,  the  notation  f(a),  /(2),  /(0),  etc.,  is  employed. 

Thus,  if  /O)  =  2  x3  -  3  x2  -  4  x  -f  5, 

then  /(*/)=  2  y»  -  3  y2  -  4  y  +  5, 

/CO  =  2  z*  -  3  z2  -  4  z  +  5, 
/(  -  x)  =  -  2  x3  -  3  x2  +  4  x  +  5, 
f{a)  =  2  a3  -  3  a2  -  4  a  +  5, 
/(2)  =  2  •  23  -  3  •  22  -  4  •  2  +  5  =  1, 
/(0)  =  5. 


244  HIGHER  ALGEBRA 

To  represent  a  function  of  several  variables,  the  notation 
f(x,  y),  f(x,  y,  z),  etc.,  is  employed.  These  are  read,  "function  of 
x  and  y,"  "  function  of  x,  y,  and  z,"  etc. 

In  the  expression  F(x,  y),  x  and  y  are  understood  to  be  inde- 
pendent of  each  other ;  but  if  the  expression  occurs  as  part  of  an 
equation,  as  f(x,  y)  =  0,  x  and  y  are  dependent.  If  the  equation 
can  be  solved  for  one  of  the  variables,  say  y,  we  shall  have 

y  =  <j>(x). 

385.  An  Increasing  Function  is  a  function  that  increases  when 
its  variable  increases,  and  decreases  when  its  variable  decreases. 

A  Decreasing  Function  is  a  function  that  decreases  when  its 
variable  increases,  and  increases  when  its  variable  decreases. 
Thus,  y  =  r3,  y  =  xh  +  x,  y  =  mx  +  b,  are  increasing  functions  ; 

y  ss  -x3,    y  =  -i   y  =  a-  x, 
x 

are  decreasing  functions  ;  while  y  =  x2  is  an  increasing  function  for  positive 

values  of  a*,  and  a  decreasing  function  for  negative  values  of  x. 

386.  A  Rational  Integral  Function  of  x  is  a  function  in  the  form 

axn  +  bx*'1  +  cxn~2  H /, 

in  which  all  the  exponents  are  positive  integers. 

387.  Functions  are  linear,  quadratic,  cubic,  etc.,  according  as 
they  are  of  the  first,  the  second,  the  third,  etc.,  degree. 

388.  The  notation  ]M  signifies  that  m  is  to  be  substituted  for 
the  variable  in  the  function  after  which  it  is  written. 

Thus,  * 3.  '        means  the  value  of  the  function  when  3  is  substi- 

xl  —  x  —  6  J3 
tuted  for  x. 

389.  A  Maximum  Value  of  a  Function  is  that  which  is  greater 
than  the  immediately  preceding  and  succeeding  values. 

A  Minimum  Value  of  a  Function  is  that  which  is  less  than  the 
immediately  preceding  and  succeeding  values. 

Thus,  for  larger  and  larger  values  of  x  in  the  function  6  x2  —  xs,  we  have 
the  following  results  : 

Values  of  a;,  -  3,  -  2,  -  1,  0,  1,    2,    3,    4,    5,  6,         7,  etc. 

Values  of /(a),      81,     32,       7,  0,  5,  16,  27,  32,  25,  0,  -  49,  etc. 


MAXIMA  AND  MINIMA   OF  FUNCTIONS  245 

It  is  seen  that  for  increasing  values  of  «,  f(x)  at  first  decreases  until  it 
reaches  0  when  x  —  0,  then  increases  until  it  reaches  32  when  x  =  4,  and 
then  decreases  for  all  larger  values  of  x.  Hence  0  is  a  minimum  value  of 
/(«),  and  32  is  a  maximum  value  of  f(x).  We  say,  also,  that  x  =  0  renders 
f{x)  a  minimum,  and  x  =  4  renders  /(x)  a  maximum. 

390.  Note.  The  greatest  value  of  a  function  is  not  necessarily 
a  maximum,  and  the  least  value  is  not  necessarily  a  minimum,  in 
the  mathematical  sense  of  these  words.  For  example,  the  least 
possible  value  of  the  function  5  +  -y/x  —  3  (using  only  the  positive 
sign  of  the  radical)  is  5,  and  this  is  when  x  =  3.  This,  however, 
is  not  a  minimum,  since  it  is  not  less  than  both  the  preceding 
and  the  succeeding  values  —  indeed,  there  is  no  preceding  value, 
for  if  x  is  made  less  than  3,  the  function  is  imaginary. 

Again,  a  minimum  of  a  function  may  be  larger  than  a  maxi- 
mum of  the  same  function.  Thus,  let  the  function  -y/x2  -f  9  be 
represented  by  y,  giving  vV  +  9  =  y,  or  x  =  ±  -y/y*  —  9.  In 
this  form  it  is  seen  that  y  can  have  no  values  between  —  3 
and  +  3,  as  these  would  render  x  imaginary,  but  can  have  all 
values  beyond  these  limits.  Hence  —  3,  being  the  largest  of  all 
the  negative  values  of  y,  is  a  maximum,  and  +3,  being  the 
smallest  of  all  the  positive  values  of  y,  is  a  minimum.  In  this 
case,  then,  the  minimum  value  of  the  function  is  larger  than  the 
maximum. 

391.  A  full  discussion  of  the  subject  of  maxima  and  minima 
of  functions  requires  the  aid  of  the  Differential  Calculus ;  but  the 
maxima  and  minima  of  many  functions  may  be  determined  by 
methods  that  are  purely  algebraic,  and  the  same  methods  apply 
to  numerous  interesting  and  important  practical  problems. 

Case  I  —  Quadratic  Functions 

392.  Theorem.  1st.  A  quadratic  function,  ax-  -f  bx  +  c,  has  a, 
maximum  if  a  is  — ,  and  a  minimum  if  a  is  +  . 

2d.    TJie  value  of  x  which  renders  the  function  a  maximum  or  a 

b 
minimum  is 

h2 

3d.    The  maximum  or  minimum  value  of  the  function  is  c 

4  a 


246  HIGHER  ALGEBRA 

Dem.     1st.   Let  ax2  +  bx  -f-  c  =  y. 
Solving  for  x,  we  have 

b        1 


x  =  —  —  ±  —  V&2  +  4  ay  —  4  ac.  (1) 

Z  a     Z  a 

If  a  is  — ,  the  term  4  ay  will  have  the  opposite  sign  from?/; 
hence  y  can  decrease  without  limit  (4  ay  becomes  +  when  y 
decreases  below  0),  but  can  increase  only  to  the  value  beyond 
which  the  quantity  under  the  radical  sign  would  become  — ,  mak- 
ing x  imaginary.     This  limit  is  a  maximum  value  of  y. 

If  a  is  +j  the  term  4  ay  will  have  the  same  sign  as  y\  hence  y 
can  increase  without  limit,  but  can  decrease  only  to  the  value 
beyond  which  the  quantity  under  the  radical  sign  would  become 
— ,  making  x  imaginary.     This  limit  is  a  minimum  value  of  y. 

2d.  In  either  case  the  limit,  which  makes  the  function  a  maxi- 
mum or  a  minimum,  is  where 

5*  +  4  ay  -  4  ac  =  0,  (2) 

which  gives  -, 

x  =  —  - — 
2a 

3d.  The  maximum  or  minimum  of  the  function  may  be  found 
either  by  substituting  this  value  of  the  variable  in  the  given  func- 
tion, or  by  solving  (2)  for  y,  which  gives 

b2 

y  =  C-  —  • 

4a 

393.  Cor.  The  value  of  the  variable  which  renders  a  pure 
quadratic  function,  ax2  +  c,  a  maximum  or  a  minimum  is  0,  and 
the  maximum  or  minimum  value  of  the  function  is  c. 

This  is  because  x  = =  —  =  0,  and  y=c—  — -  =  c  —  ----  =  c. 

2a      2a       '  *  4a  4a 

EXAMPLES   CII 

Examine  the  following  functions  for  maxima  and  minima  values : 

1.   ^-4^  +  10. 

Solution.     By  1st  and  2d  parts  of  the  theorem, 

2a  2 

renders/  (x)  a  minimum,  since  a  is  +• 


MAXIMA  AND  MINIMA    OF  FUNCTIONS  247 

By  the  3d  part  of  the  theorem, 

/  (x)  at  a  min.  =  c  -  —  =  10  -  -6  =  6. 
4  a  4 

Or  /  (x)  at  a  min.  =  x-  -  4  x  +  10]  2  =  6. 

2.  _2a2  +  6a?-5. 

Solution.  x  = = =  li 

2a  -4        2 

renders/  (x)  a  maximum,  since  a  is  — . 

/(x)  at  a  max.  =  —  2  x2  +  6  x  —  5]  ij  =  —  |. 

3.  6  a?  — a?1.  4.  »*— 8aj-hl9. 
5.  3af+12»+12.  6.  7  +  8x-2^. 

7.   5^-20^  +  20.  8.  16^-2^+18. 

9.  6«*  +  ll.  10.  -  3x2-30;  +  7. 

Case  II  —  Reciprocals  of  Quadratic  Functions 

394.  Theorem.  The  value  of  the  variable  which  renders  any 
function  a  maximum  or  a  minimum,  renders  the  reciprocal  of  that 
function  a  minimum  or  a  maximum. 

Dem.  This  is  seen  from  the  fact  that  when  f(x)  increases  or 
decreases,  ——  decreases  or  increases. 

m 

395.  Sch.  The  reciprocal  of  a  rational  integral  function  ap- 
proaches 0  as  a  limit  as  the  variable  approaches  ±  oo.  This  limit 
is  not  a  minimum  or  a  maximum  in  a  mathematical  sense  (Art. 
389),  but  is  merely  the  least  or  the  greatest  value  possible. 

EXAMPLES   CIII 

Examine  the  following  functions  for  maxima  and  minima 
values : 

1.    * 

x*-2x-\ 

Solution.     By  Art.  392, 

2  a  2 


248  HIGHER  ALGEBRA 

renders  the  denominator  a  minimum.     Hence  (Art,  394)  x  =  1  renders  the 
reciprocal  function  a  maximum. 

f(x)  at  a  max.  = =  —  -. 

1  -57 


4-6^-3^  2^  +  6^-5 

18  m  n  -?  4 


x  +  7  3  +  20x-5a? 


Case  III  —  Algebraic  Functions  of  Any  Form 

396.  Prob.  To  examine  for  maxima  or  minima  values  algebraic 
functions  of  any  form. 

Rule.  1st.  Place  the  function  equal  to  y,  and  solve  for  the 
variable  in  terms  of  y. 

2d.  If  the  result  does  not  involve  a  radical  of  even  degree,  the 
function  has  'neither  a  maximum  nor  a  minimum. 

3d.  If  the  result  involves  a  radical  of  even  degree,  place  this  radi- 
cal part  equal  to  0  and  solve  for  y.  Each  resulting  real  value  of  y 
will  be  a  maximum  or  a  minimum  according  as  an  increase  or  a 
decrease  would  give  an  imaginary  value  of  the  variable. 

Dem.  If  after  expressing  the  variable  in  terms  of  y  there  is  no 
radical  of  even  degree,  it  is  evident  that  y,  which  is  the  function, 
can  increase  or  decrease  without  limit,  every  value  of  y  giving  a 
real  value  of  the  variable. 

But  if  there  is  a  radical  of  even  degree,  and  the  quantity  under 
the  radical  sign  is  capable  of  becoming  negative  for  any  value  of  y, 
y  can  increase  or  decrease  only  to  the  limit  beyond  which  this 
quantity  would  become  negative,  making  the  variable  imaginary. 
This  limit,  which  is  reached  where  the  radical  part  equals  0,  is  a 
maximum  or  a  minimum  according  as  an  increase* or  a  decrease 
would  give  an  imaginary  result. 

397.  Sch.  1.  The  method  has  its  limitations  in  our  inability 
to  solve  equations  of  all  forms. 

398.  Sch.  2.  Case  III  includes  Cases  I  and  II,  but  the  special 
theorems  of  those  cases  give  the  result  with  less  labor. 


MAXIMA   AND  MINIMA    OF  FUNCTIONS  249 

EXAMPLES   CIV 

Examine  for  maxima  and  minima  values  the  following  func- 
tions : 

1.    a  (x  —  b)4  +  c. 

Solution.     Let  a(x  —  b)A  +  c  =  y. 

Then  (X-by  =  £^, 


y    a 


and  x  =  b  ± 


'    a 


In  this  form  it  is  seen  that  y  can  increase  without  limit,  hut  can  decrease 
only  to  y  =  e,  which  is,  therefore,  a  minimum  value.  This  makes  the  radical 
part  0,  and  leaves  x  =  b. 

2.    b  +  (x  —  a)3. 

Solution.     Let  b  +  (x  —  a)3  =  y. 

Then  x  =  a  -f  Vy  —  b. 

In  this  form  it  is  seen  that  y  can  increase  and  decrease  without  limit. 
Therefore  the  function  has  neither  a  maximum  nor  a  minimum. 

In  the  first  example,  if  y  is  made  less  than  c,  we  have  the  4th  root  of  a 
negative  quantity,  which  is  imaginary  ;  but  in  this  example  no  value  of  y 
will  give  an  imaginary  result,  as  an  odd  instead  of  an  even  root  is  involved. 

x2  -  2  x  +  13 

4  x  -  12 

Solution.    Let  s2-2:e+13 

4x-12  y 

Then  '       x*  -  (2  +  4  y)  x  =  -  12  y  -  13, 


and  x  =  I  +  2  y  ±V4  y2  —  X  y  -  12. 

By  the  3d  part  of  the  rule,  for  a  maximum  or  minimum 

4  y2  -  8  y  -  12  =  0, 

whence  y  =  3  or  —  1. 

Any  values  of  y  between  3  and  —  1  give  imaginary  values  for  x.  From  a 
large  value  y  can  diminish  to  3,  and  from  a  small  value  (a  negative  value)  y 
can  increase  to  —  1.  Hence  3  is  a  minimum  value  of  the  function  and  —  1 
is  a  maximum.     [See  Ait.  390.] 


250  HIGHER   ALGEBRA 

The  corresponding  values  of  x  are 

£  =  l  +  2?/  =  7  and  —  1. 


2x-x2      y 
2w  6 


4    3«2- 
2x- 

-6 

X2 

SOLUTION 

.     Let 

Then 

and 

3  +  y        3  +  y 

1 


V^2  +  6  y  +  18. 


3  +  2/     3  +  y 

For  a  maximum  or  a  minimum  (3d  part  of  rule), 

f  +  6  y  +  18  =  0, 

whence  y  =  — 3  ±  3V— 1. 

This  imaginary  result  shows  that  our  supposition  that  there  is  a  limit 
beyond  which  y  cannot  go  (a  limit  beyond  which  the  quantity  under  the 
radical  sign  would  become  - ,  giving  an  imaginary  value  of  x)  is  absurd 
(Art.  227).     Hence  the  function  has  neither  a  maximum  nor  a  minimum. 

.    l-3x 

5. 


x2-2x 
Solution.     Let 

Then 


x2  -  2  x      ' 
a.  =  21_-_3±iv/(2y_3)2  +  4j 

2 


As  y  can  increase  and  decrease  without  limit,  the  function  has  neither  a 
maximum  nor  a  minimum.  This  may  also  be  shown  by  placing  the  radical 
part  equal  to  0,  which  gives  an  imaginary  result. 

6.    x*-6x2-7. 

Solution.     Let  £4  —  6  x2  —  7  =  y. 


Then  x2  =  3  ±  Vy  +  16, 


and  x  =  ±  V3  ±  y/y  +  16. 

In  this  form  it  is  seen  that  y  cannot  be  less  than  —  16.  This  is,  therefore, 
a  minimum,  and  gives  x  =  ±  V3. 

Again,  when  the  minus  sign  of  Vy  +  16  is  used,  the  largest  value  of  y  is 
that  which  makes  Vy  +  16  =  3,  which  gives  y=—7,  a  maximum,  and 
x  =  0. 


MAXIMA  AND  MINIMA    OF  FUNCTIONS  251 

7.    b  +  (x  —  a)\ 
Solution.     Let  b  +  (x  —  a)^  =  y. 


Then  x  =  a  +  V(y  -  b)'2. 

By  2d  part  of  the  rule,  the  function  has  neither  a  maximum  nor  a 
minimum. 

8.   |5-|.-  9.    b  +  (x-a)i 

10.   2-^-  11.   2^». 

x2  or 


£±2*±1.  is.    (2  a* -art*. 

a?  +  2x  +  7  K  } 


12. 

14.    2  a;4-     [6x2      44.  15 


x  +  1 


16. 


x2  +  x  +  1 
(x-a)*.  17.   3^-12^-15. 


399.  If  the  maxima  and  minima  values  of  each  of  two  variables 
involved  in  an  equation  are  required,  it  is  evident  that  we  may- 
solve  for  each  of  the  variables  in  turn,  and  proceed  as  in  Case  III. 

EXAMPLES  CV 

Examine  for  maxima  and  minima  values  each  of  the  variables 
in  the  following  functions : 

1.    x*  +  8y  +  U  =  2y2  +  4:X. 
Solution.     Arranging  with  reference  to  x,  we  have 
x?  -4x  =  2y*-8y-  14, 


whence  x  =  2  ±  V2  y*  -  8  y  -  10.  (1) 

For  the  limits  of  y  we  have 

21/2-82/ -10  =  0, 
whence  y  =  5  or  —  1. 

Since  y  can  have  no  values  between  5  and  —  1,  but  can  have  all  values 
beyond  these  limits,  5  is  a  minimum  and  —  1  a  maximum  value  of  y. 

For  both  of  these  values  of  y  equation  (1)  gives  x  =  2. 

By  solving  for  y  in  terms  of  x  it  will  be  found  that  x  has  neither  a  maxi- 
mum nor  a  minimum. 

2.  x2  +  3y2  +  36  =  2x  +  12y. 

3.  x2  +  2y2  =  4:X  +  4y  +  W. 

4.  a*  +  y2  =  2y+lo. 


252  HIGHER  ALGEBRA 

400.  When  the  theory  of  maxima  and  minima  is  to  be  applied 
to  a  practical  problem,  the  first  step  is  to  obtain  an  algebraic 
expression  for  the  function  that  is  to  be  a  maximum  or  a  mini- 
mum. If  this  contains  but  one  variable,  we  proceed  as  in  one  of 
the  three  cases  explained  in  this  section.  If  it  contains  two 
variables,  one  of  them  must  be  eliminated  by  given  relations 
before  applying  the  process. 

PROBLEMS   IN  MAXIMA   AND   MINIMA  OF  FUNCTIONS 
EXAMPLES  CVI 

1.  Divide  m  into  two  such  parts  that  their  product  shall  be  a 
maximum. 

Solution.  Letting  x  and  m  —  x  be  the  two  parts,  the  function  to  be  ex- 
amined is 

x  (w  —  x)  =  -  x2  +  mx. 

By  Art.  392,  X  =  _A  =  _J?L  =  ™ 

'  -  2a  -2      2 

renders  /(a?)  a  maximum,  since  a  is  -.     Therefore  the  two  parts  are  equal. 

2.  A  farmer  having  40  rods  of  portable  fence  wishes  to  inclose 
with  it  the  largest  possible  rectangular  sheep  yard.  What  must 
be  its  dimensions  ? 

Solution.     Letting  x  and  y  represent  the  sides,  the  function  to  be  ex- 
amined is  xy.  (1) 
Since  this  contains  two  variables,  one  of  them  must  be  eliminated.     This 
is  done  by  means  of  the  condition 

2  x  +  2  y  =  40, 
whence  y  =  20  -  x.  (2) 

Substituting  this  in  (1),  the  function  to  be  examined  becomes 
x(20-x)=-  x2  +  20 x. 

By  Art.  392,  x  =  -^-  =  -  —  =  10 

J  2a  -2 

renders  f(x)  a  maximum,  since  a  is  — . 

Substituting  this  in  (2),  we  have 

y  =  20  -  x  =  20  -  10  =  10. 

Hence  the  yard  must  be  a  square. 

3.  Divide  12  into  two  such  parts  that  the  sum  of  their  squares 
shall  be  a  maximum. 


MAXIMA  AND  MINIMA    OF  FUNCTIONS  253 

4.  A  farmer  wishes  to  fence  off  a  10-acre  field  in  the  form  of 
a  rectangle  of  such  dimensions  as  to  require  the  least  amount  of 
fence.     Find  its  dimensions. 

5.  If  it  is  specified  that  a  purchaser  is  to  have  a  rectangular 
plot  of  ground  of  such  dimensions  that  3  times  its  breadth  added 
to  2  times  its  length  shall  equal  96  yards,  what  is  the  greatest 
amount  of  land  he  can  take  ? 

6.  An  aqueduct  consists  of  two  vertical  walls  surmounted  by  a 
semicylindrical  arch,  the  stone  bottom  being  of  the  same  thick- 
ness as  the  walls.  Find  the  dimensions  such  that  with  a  given 
amount  of  material  the  capacity  shall  be  a  maximum. 

Solution.  The  capacity  will  be  greatest  when  the  area  of  a  cross  section 
is  a  maximum. 

Let  x  be  the  height  of  the  side  walls,  2  y  the  breadth,  and  p  the  perimeter, 
which  is  constant,  since  the  amount  of  material  is  given.  Then  the  function 
to  be  examined  is 

o 

(1) 


(2) 


•*.+  =£ 

To  eliminate 
lence 

x,  we  have 

p  =  2x 
x=P~ 

+  2  y  +  iry, 

2  y  —  try 
2 

Substituting 

this  in  (1), 

we  have 

py-2if 

-  Try2  + 

iry2  _      4  + 
2                2 

-y2  +  . 

By  Art.  392, 

y  =  - 

b  _ 
2a 

P 

p 

-(4  +  7T) 

4  +  *- 

aders/(y)  a  maximum,  since  a  is 

-. 

Substituting 

this  in  (2), 

we  have 

2 

P     . 

4  +  7T 

irp         _ 

.     P 

2(4   +   7T) 

4  +  7T 

Hence  the  height  of  the  side  walls  equals  one  half  the  breadth  of  the  aqueduct. 

7.  A  ship  steaming  north  12  knots  an  hour  sights  another  ship 
10  knots  directly  ahead  steaming  east  9  knots  an  hour.  If  each 
keeps  on  her  course,  what  will  be  the  least  distance  between  them, 
and  at  what  time  will  it  occur  ? 

Sug.  Let  x  be  the  time,  and  find  in  terms  of  x  an  expression  for  the  square 
of  the  distance  between  the  ships.  The  distance  will  be  a  minimum  when 
the  square  of  the  distance  is  a  minimum. 


254  HIGHER   ALGEBRA 

8.  1  have  material  enough  for  a  stone  wall  48  rods  in  length. 
I  can  use  for  one  side  any  desired  portion  of  a  wall  already  built. 
What  is  the  largest  rectangular  area  I  can  inclose,  and  what  are 
its  dimensions  ? 

Show  that  the  relative  dimensions  would  be  the  same,  whatever 
the  amount  of  wall. 

9.  Find  the  side  of  the  least  square  that  can  be  inscribed  in  the 
square  whose  side  is  m.  Find  also  the  distance  of  its  corners 
from  the  corners  of  the  given  square. 

Sug.  Let  x  and  m  —  x  be  the  distances  from  a  corner  of  the  inscribed 
square  to  the  adjacent  corners  of  the  given  square.  It  will  then  be  found  that 
the  function  to  be  examined  is 

2  x2  -  2  mx  +  m2. 

10.  A  circular  piece  of  tin  is  to  be  utilized  for  the  bottom  of  the 
largest  possible  rectangular  box  of  given  depth.  Find  the  dimen- 
sions of  the  bottom. 

Solution.  As  the  depth  is  fixed,  the  volume  of  the  box  will  be  a  maximum 
when  the  area  of  the  bottom  is  a  maximum. 

Representing  by  x  and  y  the  half  sides  of  the  bottom,  the  function  to  be 
examined  is  4  xy.  Eliminating  y  by  the  relation  x2  +  y2  =  r2,  we  have  for  the 
function 


4z 

Vr2 

-X2. 

Proceeding  a 

A  in 

Case  III, 
u 

x 

Art.  i 
==  4x  ■ 

~4 

396, 

we  have 

Vr1  - 

-X2, 

V±\ 

whence 

V4r4-  w2. 

For  a  maximum  or  a  minimum 

4r4_ 

■  u2  = 

:0, 

which  gives 

u  - 

r2 
=  2' 

and 

X  - 

r 

'  V2* 

Substituting 

this  value  of 

x  in  x2 

+  y- 

■  =  r2,  we  have 

y-- 

r 

"vS* 

Therefore  the  rectangle  is  a  square. 

11.   In   the   last   example,  if  the  bottom  is  to  be  cut  from  a 
semicircle  of  radius  r,  what  must  be  its  dimensions  ? 


ZERO,   INFIX ITY,   AND  INDETERMINATE  FORMS      255 

12.  Find  the  greatest  right  triangle  that  can  be  constructed 
upon  a  given  line  as  a  hypotenuse. 

13.  A  carpenter  wishes  to  make  the  largest  possible  rectangular 
table  top  from  a  board  10  feet  long,  3  \  feet  wide  at  one  end,  and 
1  foot  wide  at  the  other.     Find  the  dimensions. 

SECTION  II  — ZERO,   INFINITY,   AND  INDETERMINATE 

FORMS 

401.  The  symbol  0  is  used  not  only  to  represent  the  absence  of 
value,  but  also  to  represent  a  quantity  that  is  less  than  any 
assignable  value ;  i.e.,  it  may  represent  either  absolute  zero  or  an 
infinitesimal.  * 

Although  all  infinitesimals  are  not  equal,  they  are  all  repre- 
sented by  the  same  symbol. 

402.  The  symbol  oo,  called  Infinity,  represents  a  quantity  that 
is  greater  than  any  assignable  value. 

Although  all  infinities  are  not  equal,  they  are  all  represented 
by  the  same  symbol. 

403.  Prob.     To  interpret  the  forms  -  and  — 

Solution.     If  in  the  fraction  -,  x  diminishes  while  a  remains 

x 

constant,  the  quotient  increases ;  and  finally  when  x  becomes  less 
than  any  assignable  value,  the  quotient  becomes  greater  than  any 

assignable  value.      Hence  ^  =  oo  • 

Again,  if  in  the  fraction  -,    x  increases  while  a  remains  con- 

x 
stant,  the  quotient  decreases ;  and  finally,  when  x  becomes  greater 
than  any  assignable  value,  the  quotient  becomes  less  than  any 

assignable  value.     Hence  —  =  0. 

00 

0  oo 

404.  Prob.     To  interpret  the  forms  -  and  — ,  called  indeterminate 

forms. 

x 
Solution.     In  the  fraction  -,  in  which  the  variables  are  inde- 

y 


256  HIGHER  ALGEBRA 

pendent  of  each  other,  if  x  and  y  diminish  until  they  become  less 

than  any  assignable  values,  giving  — ,  or  increase  until  they  become 

greater  than  any  assignable  values,  giving  — ,  the  fraction  is  inde- 
terminate, the  quotient  is  any  number  whatever,  and  the  condi- 
tions of  the  problem  from  which  it  results  are  fulfilled  for  every 
value. 

But  if  x  and  y  are  dependent  and  their  relation  is  known,  for 

x 
example,  if  y  =  mx,  giving  for  the  fraction  — -,  then  however  small 

or  however  large  x  may  become,  the  numerator  will  be  —  of  the 

1  m  0 

denominator,  and  the  quotient  will  be  —     Hence  if  the  form  - 

or  —  results  from  the  presence  in  both  terms  of  a  fraction  of  a 

factor  which  reduces  to  0  or  oo  for  a  particular  value  of  the  varia- 
ble, the  value  of  the  fraction  may  be  determined  by  dividing  out 
this  factor  before  evaluating. 


EXAMPLES  CVII 
Find  the  values  of  the  following : 
_    a*  +  5  a;  —  2" 


il- 


x*-2x  + 
Solution.     Substituting  1  for  x,  we  have  £  =  go  (Art.  403). 

ar>-5; 


2. 


x  +  6"| 
x  -  8  J2 


x>  +  2, 

Solution.     Substituting  2  for  x,  the  fraction  assumes  the  form  £.     We 
therefore  seek  for  a  factor  common  to  both  terms  of  the  fraction. 
X2  _  5  x  +  6  _  (X  -  3)  (x  -  2)  _  x  -  31  _  _  1 

+  4J2         6' 


a?_2a;  +  4~]  x  +  3      ~| 

s  +  1         •  **±*±*±11. 

x~^l      i  X              J° 

aj-r-lj-i  5x        J0 


INTERPRETATION  OF  NEGATIVE  RESULTS  257 

1  +  3 xl  t  1Q   ax~2  +  bx-1  +  cl  # 


3  +  5x]x  dx-2+ex~l+fX 

> _  a?  -  x  +  11  12        0J4-5aj8+6jB8+4aj-8 

a*-3a>  +  2   Ji"  '  Sa>*-13x8+30ai2-28a;+8- 


SECTION   III  — DISCUSSION   OF  FUNCTIONS  AND 
PROBLEMS 

405.  The  Discussion  of  a  Function  or  a  Problem  is  interpreting 
it  for  different  values  of  the  literal  quantities  which  enter  it.  If 
only  arbitrary  and  absolute  constants  enter  it,  the  discussion  is 
for  different  values  of  the  arbitrary  constants.  If  variables  and 
constants  enter  it,  the  discussion  is  for  different  values  of  the 
variables  alone. 

INTERPRETATION  OF  NEGATIVE   RESULTS 

406.  When  the  answer  or  one  of  the  answers  to  a  problem  is 
negative,  it  must  be  reckoned  in  the  opposite  direction  from  that 
assumed  as  positive  in  the  statement  of  the  problem.  For 
example,  let  it  be  required  to  find  the  time  when  A  will  be  (or 
was)  one  and  one  half  times  as  old  as  B,  A's  age  now  being  28  and 
B's  20.  Suppose  we  assume  that  this  ratio  of  their  ages  will  be 
at  some  time  in  the  future,  say  x  years  from  now.  By  the  condi- 
tions of  the  problem  we  have 

28  +  *  =  f(20  +  z), 

whence  x  =  —  4. 

As  we  assumed  time  in  the  future  +,  and  stated  our  equation 
in  accordance  with  this  assumption,  the  negative  result  shows  that 
the  time  is  in  the  past,  viz.,  4  years  ago. 

Had  we  assumed  the  event  in  the  past,  calling  past  time  +,  our 
equation  would  have  been 

28  -  x  =  f(20  -  x), 

whence  x  =  4 ; 

i.e.,  4  years  in  the  direction  we  assumed  as  positive,  or  4  years. ago. 
The  first  answer  obtained,  —  4,  is  considered  correct  in  the  alge- 
downey's  alg. — 17 


258  HIGHER  ALGEBRA 

braic  sense,  the  negative  of  future  time  being  past  time.  To 
obtain  the  correct  answer  in  an  arithmetical  sense,  the  words  will 
be  in  the  enunciation  of  the  problem  must  be  changed  to  was. 

If,  as  in  Ex.  1,  page  213,  the  nature  of  the  problem  is  such, 
either  as  enunciated  or  with  a  suitable  change  of  words,  as  was  for 
will  be,  loss  for  gain,  decrease  for  increase,  etc.,  as  not  to  permit  an 
interpretation  of  a  negative  result,  then  the  negative  answer  must 
be  rejected. 

INTERPRETATION   OF   IMAGINARY  RESULTS 

407.  As  stated  in  Art.  227,  an  imaginary  answer  shows  that, 
arithmetically,  the  conditions  of  the  problem  cannot,  even  with  a 
change  of  the  wording,  be  f  ulfilled.  For  example,  let  it  be  required 
to  divide  12  into  two  such  parts  that  the  sum  of  their  squares  shall 
be  54.  Eepresenting  the  two  parts  by  x  and  12  —  x,  we  have  by 
the  conditions 

x2  +  (12  -  xf  =  54, 

whence  x  =  6  ±  3  V—  1, 

and  12-ic  =  6  T3V^I. 

These  imaginary  results  show  that,  arithmetically,  12  cannot  be 
divided  into  two  parts  the  sum  of  whose  squares  is  54.  If  we 
apply  the  test  of  Art.  392,  we  find  that  the  sum  of  the  squares  of 
the  two  parts  is  a  minimum  when  each  of  the  two  parts  is  half  of 
the  number.  Hence  the  smallest  value  of  the  sum  of  the  squares 
is  62  -f  62  =  72.  Nevertheless,  algebraically,  the  conditions  of  the 
problem  are  fulfilled  by  these  results,  since 

the  sum  =  6  ±  3  V:=rl  +  6  T  3  V:rI  =  12, 

the  sum  of  the  squares  =  (6±  3V:rl)2  +  (6  T  S^/^lf  =  54. 

408.  In  discussing  a  function  which  has  no  connection  with 
a  problem,  the  following  features  should  usually  be  determined : 

1.  What  values  of  the  variable  (or  arbitrary  constants)  give 
but  one  value  to  the  function. 

2.  Between  what  limits  of  the  variable  the  function  has  more 
than  one  real  value,  and  whether  these  values  are  numerically 
equal  or  unequal. 


DISCUSSION   OF  FUNCTIONS  259 

3.  Between  what  limits  of  the  variable  the  function  is  imaginary. 

4.  What  value  of  the  variable  makes  the  function  a  maximum 
or  a  minimum,  and  the  value  of  the  function  at  a  maximum  or 
a  minimum. 

In  discussing  a  problem,  the  above  features  and  others,  including 
negative  and  indeterminate  results,  should  be  interpreted  with 
reference  to  the  particular  problem.  Of  course  if  the  solution 
involves  an  equation  of  only  the  first  degree,  there  could  be  no 
double,  imaginary,  maxima  or  minima  values. 

EXAMPLES  CVIII 
Discuss  the  following : 
1.   if  =  x4  —  x2. 
Discussion.     Solving  for  y,  we  have 


y  =±  xy/x'1  —  1. 

1st.   For  x  =  1  or  x  =  0,  y  has  but  one  value,  viz.,  0. 

2d.  For  all  values  of  x  <  —  1  and  >  +  1,  y  has  two  real  values,  numerically 
equal,  but  with  opposite  signs. 

3d.  For  all  values  of  x  between  —  1  and  0,  and  between  0  and  +  1,  y  is 
imaginary. 

4th.  Of  all  the  negative  values  of  x,  —  1  is  the  greatest,  a  maximum  ;  and 
of  all  positive  values  of  x,  +  1  is  the  least,  a  minimum. 

2.   2  ay2  =  x3  -f-  xf. 

Discussion.     Solving  for  y,  we  have 

v  =  ±    xV* 


V2a  —  x 

1st.   For  x  =  0,  y  has  but  one  value,  viz. ,  0. 

2d.   For  x  >  0  and  <  2  a,  y  has  two  real  values,  numerically  equal,  but 
with  opposite  signs. 

3d.    For  x  >  2  a  or  <  0,  y  is  imaginary. 

4th.   x  =  0  is  a  minimum,  and  x  =  2  a  is  a  maximum. 

5th.    For  x  =  2  a,  y  =  <x>. 

3.   3  x2  +  if  +  4  xy  -  12  x  -  8  y  +  21  =  0. 
Discussion.     Solving  for  y,  we  have 

if-  +  (4  x  -  8)  y  =  -  3  x2  +  12  x  -  21, 
or  V  =  —  2x-f4±  Vx'2  —  4  x  —  5. 


260  HIGHER   ALGEBRA 

1st.  When  x2  —  4  x  —  5  =  0,  which  gives  x  =  5  or  —  1,  y  has  but  one 
value,  viz.,  —  6  for  the  former  and  6  for  the  latter. 

2d.  When  x2  —  4  x  —  5  >0,  i.e.,  when  it  is  positive,  y  has  two  real,  un- 
equal values.     This  gives,  by  solving  the  inequality,  sc>5  or  <—  1. 

3d.  When  x2  —  4x  —  5<0,  i.e.,  when  it  is  negative,  which  gives  x<5 
and  >—  1,  y  is  imaginary. 

4th.  Since  any  values  of  x  between  —  1  and  5  give  imaginary  values  for 
y,  while  any  values  of  x  <  —  1  or  >  5  give  real  values  for  y,  —  1  is  a  maxi- 
mum value  of  x,  and  5  is  a  minimum  value  of  x. 

a'  —  a 


1  +  aa' 
Discussion.     1st.   When  a'>a,  y  is  +  ,  and  when  a'  <.a,  y  is  — . 

2d.    When  a'  =  a,  y  =  — — ,  =  0. 
1  +  a2 

a 

3d.    When  1  +  aa'  —  0,  or  a'  — ,  y  =  — rr —  =  co. 

5.   xy-  =  ±a2(2a-x).  6.    (x-l)y2  =  x2. 

7.   x2y2  =  a2  (x2  -f-  2/2).  8.   x2y  =  4  a2  (2  a  —  ?/). 

9.   if  =  a3  —  x2.  10.   ?/2  —  2  #2/  -f-  a?2  —  4  ?/  4-  x  4-  4  =  0. 

11.  Divide  a  into  two  parts  whose  product  shall  be  p. 
Solution  and  Discussion.     Representing  the  two  parts  by  x  and  a  —  x, 

we  have  x  (a  —  x)  =  p, 

whence  x  =  |  ±  \\/a2  —  4p, 

and  a  —  x  =  -  T  I  vV2  —  4p. 

m 

1st.   For  4p  =  a2,  oj  has  but  one  value,  viz.,  -. 

2i 

2d.   For  4  j9  <  a2,  x  has  two  real,  unequal  values. 

3d.   For  4p>a2,  x  is  imaginary,  showing  an  arithmetical  inconsistency, 

viz.,  naming  a  product  larger  than  any  parts  of  the  given  number  will  give. 

4th.    From  a2  —  4  p  =  0,  we  have  (Art.  393)  p  at  a  maximum  =  — ,  which 

4 
gives  x  =  -;  i.e.,  to  give  a  maximum  product  each  of  the  two  parts  must  be 

A 

half  of  the  given  number. 

12.  Two  couriers,  A  at  rate  a  and  B  at  rate  b,  are  traveling  on 
an  east  and  west  road.  At  noon  A  is  at  M  and  B  at  N,  c  miles 
apart.     Find  the  time  and  the  place  of  their  being  together. 


DISCUSSION   OF  FUNCTIONS  261 

Solution  and  Discussion.  Let  x  be  the  number  of  hours  between  noon 
and  the  instant  of  their  being  together.  Let  time  after  noon  and  distance  to 
the  east  of  M  be  regarded  as  positive.     Then 

ax  =  bx  +  c, 

whence  x  =  — - —  =  the  time  from  noon, 

a  —  b 

and  ax  =    ac    =  the  distance  from  31. 

a  —  b 

1st.     If  a,  6,  and  c  are  positive  and  a  >  6,  — - — .  and     ac     are  both  -t- , 

a  —  b  a  —  b 

showing  that  the  couriers  will  be  together  after  noon  and  to  the  east  of  31. 

The  place  is  also  east  of  N,  since     ac    >  c. 

a  —  b 
2d.  If  a,  by  and  c  are  positive  and  a<b,  — —  and  — — -  are  both  -, 

a  —  b  a-  b 

showing  that  the  couriers  were  together  before  noon  and  to  the  west  of  M. 

3d.   If  a,  b,  and  c  are  positive  and  a  —  6,  — - —  and     aG     both  become 

a  —  b  a  —  b 

oo,  showing  that  the  couriers  never  have  been  and  never  will  be  together. 
Their  rates  being  the  same,  the  distance  between  them  is  always  c. 

4th.    If  c  =  0  and  a  ^  6,  — - —  and     ac     are  both  0,  showing  that  the 
^      a—b  a—b 

couriers  are  together  at  noon,  Jfand  iV  now  coinciding. 

5th.   If  c  =  0  and  a  =  fc,  — ^—  and     ao     both  become  —    As  numerators 
a— b  a  —  b  0 

and  denominators  reduce  to  0  by  two  independent  conditions,  viz.,  c  =  0  and 
a  =  6,  the  values  of  the  fractions  are  indeterminate  (Art.  404),  and  represent 
any  values,  showing  that  the  couriers  are  together  all  the  time. 

Cth.    If  a  and  c  are  positive  and  b  negative,  the  time  becomes  — - —  and 

a  -\-b 
the  distance  — - —    Since  these  are  both  -}-,  the  couriers  will  be  together 

a  +  b 
after  noon  and  to  the  east  of  M.     The  place  is  between  M  and  N,  since 

ac    <c.    B'a  negative  rate  means  that  he  is  traveling  to  the  west. 
a  +  b 

7th.    If  b  and  c  are  positive  and  a  negative,  the  time  becomes — - 

_  —a  —  b 

and  the  distance ^— ,  or  — — —    Since  the  time  is  —  and  the  distance 

—a—b         a+b 

-f,  the  couriers  were  together  before  noon  and  to  the  east  of  31. 

8th.   If  a  and  c  are  positive  and  b  =  0,  the  time  becomes  -  and  the  dis- 

a 
tance  c.     Hence  the  couriers  will  be  together  after  noon  and  at  c  miles  east  of 
M.     The  condition  6=0  means  that  B  remains  at  N. 

9th.   If  b  and  c  are  positive  and  a  =  0,  the  time  becomes  -  and  the  distance 

b 
0.     Hence  the  couriers  were  together  before  noon  and  at  M. 
If  c  w,  re  made  negative,  it  would  locate  N  west  of  M. 


262  HIGHER  ALGEBRA 

Thus  all  the  features  of  the  problem  are  correctly  revealed  by  the  formulas 
for  the  time  and  the  distance,  the  analytic  results  agreeing  with  the  conditions. 

13.  Divide  a  into  two  parts  the  difference  of  whose  squares  shall 
be  cl. 

What  is  the  character  of  the  parts  when  d  >  a1  ?  Why  has  d  neither  a 
maximum  nor  a  minimum  value  ? 

14.  Two  couriers,  A  and  B,  are  traveling  the  same  road,  the 
former  at  rate  a,  and  the  latter  n  times  as  fast.  They  are  at  two 
places  c  miles  apart  at  the  same  time.  Find  the  time  and  the 
place  of  their  being  together. 

15.  A  cistern  can  be  filled  by  one  pipe  in  m  minutes  and  by 
another  in  n  minutes,  and  can  be  emptied  by  a  third  in  p  minutes. 
In  what  time  will  it  be  filled  if  all  are  left  open  at  once  ? 

16.  A  and  B,  traveling  the  same  road,  were  at  two  towns  m 
miles  apart  at  the  same  time.  On  coming  together  it  was  found 
that  A,  the  faster  traveler,  had  gone  n  miles,  and  that  their  time 
on  the  road  was  equal  to  the  difference  of  their  rates.  Find  their 
rates. 

17.  The  loudness  of  one  church  bell  is  m  times  that  of  another. 
If  the  amount  of  sound  varies  directly  as  the  loudness  and 
inversely  as  the  square  of  the  distance,  where  on  the  line  of 
the  two  will  the  bells  be  equally  well  heard,  the  distance  between 
them  being  a  ? 


CHAPTER  XIX 
DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS 

409.  The  Differential  of  a  Variable  or  Function  (Arts.  380  and 
382)  is  its  rate  of  change. 

If  the  change  is  uniform,  the  rate  of  change  is  the  same  as  the 
actual  change  during  the  unit  of  time;  but  if  the  change  is  not 
uniform,  the  rate  of  change  at  any  instant  is  the  change  that 
would  take  place  during  the  next  unit  of  time  if  it  continued 
uniform  at  what  it  is  at  that  instant. 

Thus,  we  say  that  a  body  falling  from  rest  has  at  the  end  of  the  first 
second  a  velocity  of  32£  feet.  By  this  we  do  not  mean  that  it  falls  only  32£ 
feet  during  the  next  second,  but  that  if  it  continued  to  fall  uniformly  with  the 
velocity  it  has  at  that  instant,  it  would  fall  32£  feet  during  the  next  second. 
The  distance  from  the  falling  point  is  variable  and  the  differential  of  that 
variable  at  the  end  of  the  first  second  is  32£  feet. 

410.  The  differential  of  a  variable  is  indicated  by  writing  d 
before  it.  Thus,  the  differential  of  x  is  written  dx,  and  is  read 
"  differential  a,*." 

411.  If  a  variable  or  function  is  increasing,  its  differential  is 
positive ;  if  decreasing,  negative. 

412.  Theorem.     Constant  terms  disappear  in  differentiating. 

For  since  a  constant  admits  of  no  change,  the  differential  of 
a  constant  is  0. 

413.  Theorem.  The  differential  of  the  product  of  a  constant  and 
a  variable  is  the  constant  multiplied  by  the  differential  of  the  variable. 

Dem.     Let  the  function  be  ax.  (1) 

Whether  x  changes  uniformly  or  not,  let  dx  be  the  amount  by 

which  it  ivoidd  increase  in  a  unit  of  time  if,  at  any  instant,  the 

change  should  become  uniform.     Then  the  state  of  the  function 

at  the  end  of  the  unit  of  time  would  be 

a(x  +  dx)  =  ax  -f  adx.  (2) 

263 


264  HIGHER  ALGEBRA 

The  difference  between  the  state  of  the  function  at  the  begin- 
ning of  a  unit  of  time  and  what,  with  a  uniform  change,  it  would 
be  at  the  end  of  the  unit  of  time  is  the  differential  of  the  function 
(Art.  409).     Hence,  subtracting  (1)  from  (2),  we  have 

d(ax)  =  adx. 

414.  Theorem.  The  differential  of  a  polynomial  is  the  algebraic 
sum  of  the  differentials  of  its  terms. 

Dem.     Let  the  function  be  x  -f  y  —  z.  (1) 

Whether  x,  y,  and  z  change  uniformly  or  not,  let  dx,  dy,  and  dz 

be  the  respective  amounts  by  which  they  would  increase  in  a  unit 

of  time  if,  at  any  instant,  the  change  should  become  uniform. 

Then  the  state  of  the  function  at  the  end  of  the  unit  of  time 

would  be 

x  +dx  +  y  +  dy-(z  +  dz).  (2) 

The  difference  between  the  state  of  a  function  at  the  beginning 
of  a  unit  of  time  and  what,  with  a  uniform  change,  it  would  be 
at  the  end  of  the  unit  of  time  is  the  differential  of  the  function 
(Art.  409).     Hence,  subtracting  (1)  from  (2),  we  have 

d (x  +  y  —  z)  =  dx  -f  dy  —  dz. 

415.  Theorem.  The  differential  of  the  product  of  two  variables  is 
the  sum  of  the  products  of  each  into  the  differential  of  the  other. 

Dem.     Let  the  function  be  xy. 

Whether  x  and  y  change  uniformly  or  not,  let  dx  and  dy  be  the 
respective  amounts  by  which  they  would  increase  in  a  unit  of 
time  if,  at  any  instant,  the  changes  should  become  uniform. 

Now  if  x  alone  were  to  change,  the  change  in  the  product  would 
be  ydx  (Art.  413),  and  if  y  alone  were  to  change,  the  change  in  the 
product  would  be  xdy  (Art.  413),  while  if  both  x  and  y  change, 
the  change  in  the  product  at  a  uniform  rate  would  be  the  sum  of 
the  changes  due  to  these  two  causes,  or  ydx  +  xdy.     Hence, 

d(xy)  =  ydx 4-  xdy* 

*  While  this  is  not  a  rigorous  demonstration,  it  is  sufficient  to  show  the 
truth  of  the  theorem. 


DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS      265 

416.  Theorem.  TJie  differential  of  the  product  of  several  variables 
is  the  sum  of  the  products  of  the  differential  of  each  into  all  the  others. 

Dem.     Let  the  function  be  xyz. 
Let  v  =xy.     Then  xyz  =  vz. 

By  Art.  415,        d(vz)  =  zdv  -f  vdz,  (1) 

and  dv  =  ydx  -f  xdy.  (2) 

Substituting  in  (1)  the  values  of  v  and  dv,  we  have 

d(xyz)  =  z(i/c?;c  -|-  ax%)  -f  #ydz, 

=  yzda?  -f  xzdy  -f  xydz. 

The  same  method  of  reasoning  will  apply  to  any  number  of 
variables. 

417.  Theorem.  The  differential  of  a  fraction  ivith  variable  nu- 
merator and  denominator  is  the  denominator  into  the  differential  of 
the  numerator  minus  the  numerator  into  the  differential  of  the 
denominator,  divided  by  the  square  of  the  denominator. 

Dem.     Let  the  function  be  -,  and  represent  this  function  by  u, 

giving  y 

x 
u  =  — 

y 

Clearing  of  fractions, 

yu  =  x. 

Pifferentiating  by  Art.  415, 

ydu  +  udy  =  dx, 

.     ,   dx dy 

,        dx  —  udy              y          ydx  —  xdy 
whence  du  = -  = —  =  - ^ — -• 

y  y  y 

418.  Cor.  i.  Tfie  differential  of  a  fraction  whose  numerator  is 
constant  is  minus  the  numerator  into  the  differential  of  the  denomi- 
nator, divided  by  the  square  of  the  denominator. 

For  if  x  =  a,  a  constant,  then  da  =  0,  and  dl -)= ~- 

\yj        y 


266  HIGHER  ALGEBRA 

419.  Cor.  2.  The  differential  of  a  fraction  ivith  a  constant 
denominator  is  the  differential  of  the  numerator  divided  by  the 
denominator. 

For  -  =  -x,  and  -  is  a  constant  factor:  therefore,  by  Art.  413, 
b      b   '         b  J 

d(  -)=:d(  -x\=-dx. 


b   J     b 


b 

420.  Theorem.  The  differential  of  a  variable  having  a  cojxstant 
exponent  is  the  product  of  the  exponent,  the  variable  with  its  expo- 
nent diminished  by  one,  and  the  differential  of  the  variable. 

Dem.     Let  the  function  be  xn. 

1st.     When  n  is  a  positive  integer. 

Now  xn  =  x-x*  X'"to  n  factors. 

Hence,  by  Art.  416, 

d(xn)  =  (x  •  x  •  •  •  to  n  —  1  factors)  dx 
+  (x  •  x  •  •  •  to  n  —  1  factors)  dx  -\ to  n  terms, 

=  xn~l  dx  +  a;"-1  dx-\ to  n  terms, 

=  nxn~x  dx. 
2d.     When  n  is  a  positive  fraction,  as  — 


Let 

p 
y  =  x«. 

By  involution, 

yq  =  xp. 

By  1st  case, 

qy*-1  dy  =  pxp~x  dx, 

,            ,       px*'1  dx     px*-1  dx      p   p-\ 
whence  dy  =  i —  =- =  —  x9    dx. 

3d.    When  n  is  negative. 

Now  x~n  ==  — 

Differentiating  by  Art  418  and  1st  case, 


(,-)=^(i): 


DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS      267 

421.  Cor.  The  differential  of  the  square  root  of  a  variable  is  the 
differential  of  the  variable  divided  by  twice  the  square  root  of  the 
variable. 

For  d  hjx)  =  d(xh=l  af*  dx  =  -i^L. 

EXAMPLES  CIX 

Differentiate  the  following : 

1.  y  =  2x3-4:x2  +  5x. 
Operation.     By  Art.  414, 

dy  =  d(2  x3)  -  d(4  x2)  +  d(5  x) 

=  3-2  xHx  -  2  •  4  xdx  +  5  dx,  by  Arts.  413  and  420. 
=  (6x2-8x  +  5)dx. 

This  means  that  y,  or  2  x8  —  4  x2  -f  5  x,  -changes  6  x2  —  8  a;  +  5  times  as 
fast  as  x.     Hence, 

when  x  =  0,  y  changes  5  times  as  fast  as  x  ; 
when  x  =  1,  y  changes  3  times  as  fast  as  x  ; 
when  x  =  2,  y  changes  13  times  as  fast  as  x  ; 
when  x  =  3,  y  changes  35  times  as  fast  as  x. 

2.  y  =  x?  +  5x*-Q>x  +  T.  3.   y  =  x-s-4x2  +  25. 

4.  y  =  7x4  +  3x2-6xl.  5.   ?/  =  2^  -  oar4  -  5r*  +  9. 

6.  ?/  =  3(5-f-4ar})5. 

Operation.     Treating  the  part  in  the  parenthesis  as  the  variable,  we  have 
dy  =  5  .  3(5  +  4  x8)*3  •  4  x2dx  =  180(5  +  4  x3)Wx. 

7.  y  =  5  (2  +  3  a,-2)4.  8.   y  =  4  (1  -  5  a?2)3. 
9.   ?y  =  (a  +  ^)i                            10.   2/  =  2(3-arJ)"2. 

11.  7*  as  X2!/*. 

Operation.     By  Art.  415, 

du  =  y3<7(x2)  +x*d(y*)  =  2  xy3dx  +  3  x2J/2(7y. 

12.  w  =  afy*.  13.    w  =  ar(.y2-3). 
14.   u =  (x  —  1)  (f  +  2).  15.   M  =  aj(a?-|-a)i 

16.    2/=— t— 


268  HIGHER  ALGEBRA 

Operation.     By  Art.  417, 

.    _  x*d(x*  -  3)  -  O2  -  3)d(x3) 


y  — 

2  a;4^  _ 

-  3  z4<fcc  +  9  xHx  _  9  -  z2^ 

17     V  — 

421, 

X?                                        X* 

18.    tt  =  -. 

2/2 

'   f      (1  +  a^)3 

19.  y=V3H-5#. 

Operation.     By  Art. 

• 

2a/3  +  5x3      2V3  +  5x: 


20.    ?/  =  VI  +  a?.  21.    y  =  4 V2  x  —  3  3. 

422.  An  Equicrescent  Variable  is  one  which  changes  uniformly, 
i.e.,  one  which  has  a  constant  rate. 

423.  A  Second  Differential  is  a  differential  of  a  differential,  i.e., 
the  rate  at  which  the  rate  is  changing. 

The  differential  of  dy  is  written  d2y,  and  is  read,  "  second  dif- 
ferential y  " ;  while  dx  x  dx,  or  the  square  of  dx,  is  written  dx2. 

From  the  function  y  =  ax3, 

we  have  dy  =  3  ax^dx  =  (3  adit)  x2. 

If  we  regard  x  as  an  equicrescent  variable,  dx  is  constant ;  but 
dy  is  variable,  since  it  is  equal  to  an  expression  containing  a  vari- 
able x. 

Hence  we  have 

d  (dy)  =  d2y  =  d  [(3  adx)  x2^  =  3  adx  x  2  #cZ#  =  6  axdx2. 

424.  A  Third  Differential  is  the  differential  of  the  second  dif- 
ferential, and  so  on. 

425.  The   First  Derivative,  or   the   First  Differential  Coefficient, 

is  the  ratio  of  the  differential  of  the  function  to  the  differential 

of  the  variable,  and  is  represented  by  — ,  or  f'(x). 

(XX 


DIFFERENTIATION  OF  ALGEBRAIC  FUNCTIONS      269 

Thus,  from  y  =  x:i  +  3  x2  —  5  x, 

we  have  dy  =  (3  x2  +  6  x  —  5)dx, 

and  ^  =  3  x2  +  6  x  -  5, 

dx 

in  which  3  x2  +  6  x  —  5  is  the  first  derivative. 
The  same  thing  is  expressed  by  writing 

/(x)  =  x3  +  3  x2  -  5  x, 

and  f'(x)  =  3  x2  +  6  x  -  5. 

426.  The  Second  Derivative,  or  the  Second  Differential  Coefficient,  is 
the  ratio  of  the  second  differential  of  the  function  to  the  square  of 

the  differential  of  the  variable,  and  is  represented  by  — %  or  f"(x). 

427.  The  Third  Derivative,  or  the  Third  Differential  Coefficient,  is 
the  ratio  of  the  third  differential  of  the  function  to  the  cube  of 

d3y 
the  differential  of  the  variable,  and  is   represented  by   — ^,  or 

f"\x),  and  so  on.  d^ 

EXAMPLES  CX 

Find  the  first  derivative  of  each  of  the  following : 

1.  ?/  =  2  +  3#-4arJ. 

Operation.  dy  =  3  dx  -  12  x2dx, 

whence  ^  =  3  -  12  x2. 

dx 

2.  ?/  =  ar5-3ar2  +  4.  3.   y  =  (x3  -  5)3. 

4.   y  =  Var*  —  2.  5.   y  =  xr  -f  x  -f  ar3. 

6.   y  =        m      •  7.    y=        m      ■ 

(1  +  *)2  F      (*  +  a)4 

Find  the  second  derivative  of  each  of  the  following : 

a   2/  =  o(3  +  4z)3. 
Operation.  /'(x)  =  60(3  +  4  x)2, 

and  /"(x)  =  480(3  +  4x). 

9.  y  =  x2-6x  +  7.  10.  2/  =  3»4-5^  +  10. 

11.   y  =  3(2  +  ox)\  12.   ?/  =  (a  +  z)-3. 

13.  y  =*_**-.  14.  y  =  5(2x2-3)4. 


270  HIGHER  ALGEBRA 

Find  the  first  four  successive  derivatives  of  each  of  the 
following : 

15.   y  =  (a  +  x)-2.  16.   y  =  ^-- 

a-\-  x 

17.   y  =  3(l  +  xy.  18.   y  =  3xi-lxi+3x2  +  6x-7. 

19.  y  =  2(5  —  x)K 

20.  y  =  A  +  Bx  +  Car9  -f-  Dar3  +  j£af\ 

428.  A  Partial  Differential  of  a  function  of  more  than  one  variable 
is  its  differential  on  the  hypothesis  that  all  the  variables  but  one 
become  constant. 

Thus,  from  u  =  3  xy  +  5  x  —  4  y  +  7, 

we  have  dxu  =  3  y&c  -f  5  dx, 

and  c^w  =  3  xdy  —  4  dy, 

the  subscripts  indicating  with  reference  to  which  variable  we  have  differ- 
entiated. 

429.  A  Partial  Derivative,  or  a  Partial  Differential  Coefficient,  is 
the  ratio  of  the  partial  differential  of  the  function  to  the  dif- 
ferential of  that  variable  with  reference  to  which  we  have 
differentiated. 

Thus,  from  the  above  function  we  have 

—  =  3  y  +  5, 
dx 

and  ^  =  3  x  -  4, 

dy 

the  denominators  indicating  with  reference  to  which  variable  we  have  dif- 
ferentiated. 

430.  Theorem.  In  an  algebraic  function  of  the  sum  of  two  vari- 
ables, x  and  y,  the  first,  second,  third,  etc.,  partial  derivatives  with 
reference  to  x  are  equal  respectively  to  the  first,  second,  third,  etc., 
partial  derivatives  ivith  reference  to  y.* 

*  This  is  true  for  any  function  of  the  sum  of  two  variables,  but, it  is 
proved  here  only  for  algebraic  functions. 


DIFFERENTIATION   OF  ALGEBRAIC  FUNCTIONS      271 

Dem.  All  algebraic  functions  of  the  sum  of  two  variables  are 
included  in  the  form 

u  =  m(x  +  y)n, 

in  which  m  and  n  are  positive  or  negative,  integral  or  fractional. 

Now  dxu  —  mn  (x  +  y)n~l  dx, 

and  dyii  =  mn  (x  +  y)n~l  dy, 

which  forms  differ  only  in  the  factors  dx  and  dy ;  but  in  forming 
the  derivatives  these  factors  are  divided  out,  giving 

da      du  /     .     n„_, 

—  =  —  =  mn  (x  -f  «)     . 
dx     dy  K    ^y) 

This  is  a  new  function  of  the  sum  of  two  variables,  and,  as 
just  shown,  its  first  partial  derivatives  are  equal;  but  these  are 
the  second  partial  derivatives  of  the  original  function,  giving 


Similarly, 


431.  Let  the  student  form  the  first  three  successive  partial 
derivatives  of  u  =  2  (x  +  y)z,  u  =  6  x4  +  3  y3,  u  =  5  (x  -f-  y)~'2, 
u  =  -yjx  +  y,  u  =  5  (x  —  y)7,  u  =  (x2  +  3  y)4,  u  =  (x  -f  y) *,  and 
observe  that  the  law  stated  in  the  theorem  holds  for  all  straight 
functions  of  the  sum  of  two  variables,  while  it  does  not  for  other 
functions ;  though  in  a  straight  function  of  the  difference  of  two 
variables  the  corresponding  partial  derivatives  are  equal  except 
in  sign. 


dru  _  dru 
dx2     dy2 

d?u  _  cPu    d*u  _  d4u 
dx3     dy*    dx*     dy4 

dnu  _  dnu 
dxn      dyn 

CHAPTER   XX 
DEVELOPMENT  OF  FUNCTIONS 

432.  A  Finite  Series  is  a  series  (Art.  312)  which  terminates. 

433.  An  Infinite  Series  is  a  series  which  does  not  terminate,  but 
has  an  endless  succession  of  terms. 

434.  An  infinite  series  is  Convergent  when,  by  taking  more  and 
more  terms,  the  sum  approaches  a  finite  limit;  it  is  Divergent 
when  the  sum  does  not  approach  a  finite  limit ;  and  it  is  Oscillating, 
or  Neutral,  when  the  sum,  though  finite,  does  not  approach  a 
determinate  limit. 

Thus,  1  +  x  +  x2  +  x3  +  x*  -f  x5  +  ••• 

is  convergent  when  x  is  numerically  less  than  1,  approaching  the  limit  of 

(Art.  329)  ;  divergent  when  x  is  1  or  greater  than  1  ;  oscillating  when  x  is  —  1, 
the  sum  being  0  or  1  according  as  the  number  of  terms  is  even  or  odd. 

435.  Note.  Convergent,  when  applied  to  a  series,  is  not  synony- 
mous with  decreasing  or  descending  (Art.  313).  While  every 
convergent  series  is  decreasing,  not  every  decreasing  series  is  con- 
vergent. Thus,  1,  i,  J,  J,  i,  •••  is  decreasing,  but  is  not  convergent, 
as  shown  in  Art.  583  (3). 

436.  To  Develop  or  Expand  a  Function  is  to  find  a  series  whose 
sum  is  equal  to  the  function.  The  value  of  a  developed  function, 
therefore,  is  either  the  sum  of  a  finite  series,  or  the  limit  of  the 
sum  of  an  infinite  convergent  series. 

437.  There  are  many  ways  of  developing  functions.  The  stu- 
dent is  already  familiar  with  development  by  division,  involution, 
and  evolution. 

Thus,  by  actual  division, 

— —  =1  -x  +  x^^  +  ^-x6, 
1  4-  x 

272 


DEVELOPMENT  OF  FUNCTIONS  273 

a  finite  series  and  equal  to  the  function  for  all  values  of  x  ;  by  the  binomial 
theorem  (Art.  155), 

(1  +  x)5  =  1  +  5  x  +  10a;2  +  10  x*  +  5x*  +  x5, 

a  finite  series  and  equal  to  the  function  for  all  values  of  x  ;  and  by  extracting 
the  square  root  of  1  —  x2  we  have 

vrr^i-^-*4-*  -..., 

2       8       lo 

an  infinite  series  and  true  only  for  values  of  x  numerically  equal  to  or  less 
than  1,  being  divergent  for  all  other  values. 

438.   It  often  occurs,  as  in  the  last  example,  that  a  development 
is  true  for  certain  values  of  the  variable,  but-  not  for  all  values. 

Thus,  by  division, 

=  l+z  +  £2+a;3  +  £4  +  .... 


1-x 

Now  when  x  is  a  proper  fraction,  either  positive  or  negative,  the  series  is 
convergent,  and,  if  we  treat  it  as  an  infinite,  decreasing  geometrical  progres- 
sion, the  formula 

S  =  -2-  (Art.  329)  gives 


1  -  r  '  "        1  -  x 

and  the  series  is  seen  to  be  equal  to  the  function. 

When  x  is  greater  than  1,  the  function  (the  first  member)  is  a  negative 
fraction,  while  the  series  (the  second  member)  is  positive  and  departs  farther 
and  farther  from  the  value  of  the  function  as  more  and  more  terms  are  taken. 

When  x  is  numerically  greater  than  1,  but  negative,  the  function  is  a  posi- 
tive fraction,  while  the  series  is  positive  or  negative  according  as  the  number 
of  terms  is  odd  or  even,  and  departs  farther  and  farther  from  the  value  of 
the  function  as  more  and  more  terms  are  taken. 

When  35=— 1,  the  function  is  £,  while  the  series  is  1  —  1  +  1  —  1  +  1-1+  •••, 
which  is  0  or  1  according  as  the  number  of  terms  is  even  or  odd. 

When  as.ss  1,  both  the  function  and  the  series  become  oo. 

If  the  order  of  the  terms  in  the  divisor  be  reversed,  we  have 

1  1111 


-  X  +  1  X        X2       Xs       X* 

and  the  series  equals  the  function  for  values  of  x  numerically  greater  than  1, 
but  not  for  values  numerically  less  than  1. 

439.    The  subject  of  Convergence/  is  reserved  for  a  subsequent 
part  of  the  work,  and  we  shall  in  this  chapter  concern  ourselves 
only  with  the  Development. 
Downey's  alg. — 18 


274  HIGHER  ALGEBRA 

SECTION  I  — INDETERMINATE   COEFFICIENTS 

440.  Indeterminate  (or  Undetermined)  Coefficients  are  assumed 
constant  coefficients  whose  values,  not  known  at  the  outset,  are 
to  be  determined  in  the  course  of  the  demonstration  of  a  theorem 
or  the  solution  of  a  problem. 

441.  Theorem  of  Indeterminate  Coefficients.  If  tico  series  con- 
taining the  same  variable  are  equal  for  all  values  of  the  variable  that 
render  both  series  convergent,  the  coefficients  of  the  same  poivers  of 
the  variable  in  the  two  series  are  equal. 

Dem.     Let 

A  +  Bx  +  Cx2  -f  Dx>  +  etc.  =  A'  -f  B'x  +  C'x2  +  D'x*  +  etc. 

for  all  values  of  x  which  render  each  series  either  finite  or  infinite 
and  convergent ;  then  A  =  A ',  B  =  B',  C=  C,  D  =  I)',  etc. 

Since  the  equation  is  true  for  all  values  of  x  which  render  each 
series  either  finite  or  infinite  and  convergent,  it  is  true  when 
x  =  0.  But  this  gives  A  =  A';  and  since  A  and  A'  are  constant, 
they  have  the  same  values,  whatever  the  value  assigned  to  x,  so 
long  as  each  series  is  either  finite  or  infinite  and  convergent. 

Dropping  these  equals  from  the  two  members  and  dividing  by  x, 

B+Cx  +  Dx2  +  etc.  =  B'  +  C'x  +  D'x2  +  etc. 

As  before,  making  x  =  0,  we  have  B  =  B'.  Proceeding  in  the 
same  way,  we  have  C  =  C,  D  =  D\  etc. 

Cor.  If  A  -f  Bx  +  Cx2  -f  Dx3  -f  etc.  =  0  for  all  values  of  x, 
each  of  the  coefficients  A,  B,  C,  etc.,  is  0. 

For  we  may  write 

A  +  Bx  +  Cx2  +  Dx3  +  etc  =  0  +  0^  +  0^  +  0.^  +  etc., 
whence,  by  the  theorem,  A  —  0,  B  =  0,  (7=0,  etc. 

442.  The  theorem  of  Art.  441  gives  a  method,  called  the  Method 
of  Indeterminate  (or  Undetermined)  Coefficients,  of  developing 
a  function  into  a  series.  It  consists  in  assuming  a  series  in 
ascending  powers  of  the  variable  with  unknown  coefficients,  and 
then  finding  the  values  of  these  coefficients  by  equating  those  of 
the  same  powers  of  the  variable. 


INDETERMINATE  COEFFICIENTS 


275 


Thus,  let  it  be  required  to  develop 


2x 


Solution.    Assume 
l-3x-a 


1  -  2  x  -  x* 
Clearing  of  fractions, 

1-%X-X2  =  A  +    j5 
-2A 


=  A  +  Bx  +  Cx'2  +  Dx*  +  .ETx4  +  etc. 


-2B 

-    A 


x2+  D 
-20 
-     B 


£3+      # 

-2D 

-     C 


x*  +  etc. 


Equating  the  coefficients  of  the  same  powers  of  x  (Art.  441), 
A  =  l; 

B-2A  =  -3,  whence  B  =  -3  +  2A  =  -l; 
C-2B-  A  =  -l,  whence  C  =  -l  +  2B  +  A=-2 
D  -  2  C  —  B  =  0,  whence  D  =  2C+B  =  -b; 
E-2D-C  =  0,  whence  E  =  2D+  C  =  -12. 

Substituting  these  values  in  the  assumed  series,  we  have 


l-3x 


=  1  -X-2X*-  5x3-12xl 


etc. 


1  _  2  x  -  x2 
This  can  readily  be  verified  by  actual  division. 

443.  In  developing  by  the  Method  of  Indeterminate  Coefficients, 
the  Law  of  the  Series  becomes  evident  early  in  the  progress  of  the 
work,  and  then  as  many  more  coefficients  as  desired  may  be 
obtained  by  simply  combining,  according  to  the  discovered  law, 
the  coefficients  already  found. 

Thus,  in  the  example  of  the  last  article,  we  have 

D  =  2C+B, 

E  =  2  D  +  C. 
It  is  therefore  evident  that 

F  =  2  E  +  D, 

G  =  2F+  E, 

and  so  on,  each  coefficient  after  the  third  being  twice  the  preceding  plus  the 
second  preceding. 

In  all  cases  in  developing  rational  fractions  the  law  of  the  series 
appears  in  all  the  equations  beyond  the  one  ichose  second  member  is 
the  coefficient  of  the  highest  power  of  the  variable  in  the  numerator 


276  HIGHER  ALGEBRA 

of  the  given  fraction;  for  beyond  this  the  first  members  all  have 
the  same  form,  and  the  second  members  are  all  0. 

The  student  should  in  each  example  note  the  law  of  the  series. 

EXAMPLES  CXI 

Expand  by  the  Method  of  Indeterminate  Coefficients  the 
following : 

t      l+2a;  2      2-3s  3       1+3 

1  —  X  —  X2  1  -f  cs  +  x2 

4.    l±* 5.         5  +  2x      .  6. 

l  +  2x-\-3x2  1-5X  +  X2 

2-Sx2-hxi  1  +  x  +  x2 

'     % _  x 1 4.  as?  "       1  +  ar*  "  2  -  a  -  a2 

444.  In  developing  a  rational  fraction  we  must  assume  a  series 
such  that,  after  clearing  of  fractions,  the  second  member  shall 
contain  all  the  powers  of  the  variable  that  are  found  in  the  first 
member,  which  is  the  numerator  of  the  given  fraction  Other- 
wise it  would  be  necessary  to  equate  given  finite  coefficients  with 
0,  which  would  be  absurd.  The  exponent  with  which  to  begin 
the  assumed  series  may  be  determined  by  noting  what  actual 
division  would  give  for  the  first,  or  lowest,  exponent. 

1  —  x 
In  developing  — —  we  assume 

—  X    -f-  O  #/ 

Ax~2  +  3x^+0  +  Dx  +  Ex2  +  Fx3  +  etc. ; 


1-x+x2 

l  +  2x 

1-3^ 

\  +  2x 

2^  +  3^ 


1/  1  —  x 
or,  if  we  prefer,  we  may  write  the  expression  in  the  form  —  f  - — — 

X" \li  -\~  o  X J 

expand  the  part  within  the  parenthesis  by  the  use  of  the  series 
beginning  with  an  absolute  term,  and  then  multiply  each  term 

If  in  any  case  we  inadvertently  assume  a  wrong  series,  the 
fact  will  appear  by  the  absurd  results  obtained  when  the  coeffi- 
cients of  the  same  powers  of  the  variable  are  equated.  If  in  this 
example  we  should  assume  the  series 

A  +  Bx  4-  Cx2  +  Dx*  +  ExA  +  etc., 


INDETERMINATE  COEFFICIENTS  277 

clear  of  fractions,  and  equate  the  coefficients  of  the  same  powers 

of  x,  we  would  have  1  =  0,  —1  =  0,  etc.,  which  are  absurd  results. 

2  x2  —  3  Xs 

In  expanding  such  a  fraction  as  — ■ — ,  no  absurdity  would 

J.  —  x 

result  in  assuming  the  series  A  +  Bx  +  Cx2  +  Dot?  +  Ex4  -f  etc. ; 
but,  inasmuch  as  A  and  2?  would  in  that  case  be  equated  with  0, 
it  is  shorter  to  assume  the  series  Ax2  +  Bx3  -f  Cx4  +  Dx5  -f  etc. ; 
or,   if  we   prefer,   we   may   write   the   expression   in  the   form 

— — - ),  expand  the  part  within  the  parenthesis  by  the  use 
x  —  x  j 

of  the  usual  series,  and  then  multiply  each  term  by  x2. 

EXAMPLES   CXII 

Expand  by  the  Method  of  Indeterminate  Coefficients  the  fol- 
lowing : 

2  2       l  +  x-x2  „        2a? 


•   3^-4^  x-2x2  +  3x3  3-20T2 

1-2^-^  _    x-Srf-x*  c      3-2x  +  x* 

4.    5. 


a^  +  ^-x4  \-2x-x2  2x3-x4-2x6 

445.  The  Method  of  Indeterminate  Coefficients  may  also  be 
employed  for  expanding  a  radical.  If  the  quantity  under  the 
radical  sign  is  a  perfect  power  of  the  degree  of  the  index,  the 
expansion  will  give  the  root.  If  the  quantity  is  a  binomial, 
the  binomial  theorem  gives  a  more  expeditious  method  of  expan- 
sion. 

EXAMPLES  CXIII 

Expand  by  the  Method  of  Indeterminate  Coefficients  the 
following : 

L    Vl-2a;  +  3^. 
Solution.     Assume 


Vl  -  2  x  +  3  x*  =  A  +  Bx  +  Cx2  +  Dx*  +  Ex4  +  etc. 
Squaring  both  members,  we  have 
l-2x  +  S  x2  =  A2 +  2  ABx  +  B2       x2  +  2  AD  x?  +  C2      x*  +  etc. 


x  +  B2 

x2  +  2AD 

2?+   C2 

+  2  AC 

+  2BC 

+  2AE 
+  2BD 

278  HIGHER  ALGEBRA 

Equating  the  coefficients  of  the  same  powers  of  x  (Art.  441), 
A2  =  1,  whence  ^4  =  1  (using  only  the  plus  sign)  ; 

2  AB  =  -  2,  whence  B  =  —  =  -  1  ; 
•  A 

2  AC  +  B2  =  3,  whence  (7  =  8  ~  B*  =  1 ; 

2^4 

~RC1 

2AD+2BC  =  0,  whence  2)  =  -  —  =  1 ; 

^4 

2  JE1  +  C2  +  2  BD  =  0,  whence  #  =  ~  C'2  ~  2  ^  =  -» 

2.4  2 

Substituting  these  values  in  the  assumed  series,  we  have 


VT  —  2  x  +  3  x2  s=  1  —  *  -r  -*r  -r  ■*>-  - 
This  can  readily  be  verified  by  extracting  in  the  usual  way  the  square  root 
of  1  -  2  x  +  3  x2. 


2.    Vl  +  x  +  jc2.  3.    Vl  +  x  —  x1.  4.    V9  -f-  x  —  3  x2. 


5.    V4-4a  +  13a2-6ar5  +  9a;4.  6.    VI  +  *  +«*. 

DECOMPOSITION   OF   FRACTIONS 

446.  We  have  seen  (Art.  136)  that  several  fractions  may  be 
united  into  one  fraction  whose  denominator  is  the  lowest  common 
multiple  of  all  the  denominators.  The  converse  operation  of 
separating  a  fraction  into  partial  fractions  is  sometimes  neces- 
sary, especially  in  the  operations  of  the  Integral  Calculus.  The 
Method  of  Indeterminate  Coefficients  furnishes  a  means  of  mak- 
ing this  separation. 

If  the  numerator  of  the  fraction  to  be  separated  is  of  higher 
degree  than  the  denominator,  it  should,  by  division,  be  made 
lower.     Thus, 

2ss  +  3s2-4a;-f  2  =  0  .    8x-5     § 

0?--2*;+l  V-2*+l" 

Case  I 

447.  TT/ten  ^e  denominator  is  resolvable  into  equal  or  unequal 
factors  of  the  first  degree. 

Rule.  Assume  the  given  fraction  equal  to  several  fractioyis  with 
undetermined  numerators,  and  whose  denominators  are  all  of  the 
divisors  of  the  denominator  of  the  given  fraction. 


DECOMPOSITION   OF  FRACTIONS  279 

Clear  the  equation  of  fractions  and  collect  terms. 
Equate  the  coefficients  of  the  same  powers  of  the  variable,  and 
determine  the  values  of  the  assumed  numerators. 
Substitute  these  values  in  the  assumed  fractions. 

Dem.  Since  the  given  fraction  is  the  sum  of  the  partial  frac- 
tions, each  denominator  must  be  a  divisor  of  the  given  denomi- 
nator; and  since  any  divisor  is  a  possible  denominator,  all  the 
divisors  must  be  used. 

The  numerator  of  an  assumed  fraction  must  contain  only  the 
undetermined   constant,   since   otherwise   the    assumed    fraction 

Aw  J\a 

would  be  capable  of  farther  separation  ;  thus,  =  A  + 


x  —  a  x  —  a 

After  clearing  the  equation  of  fractions,  the  coefficients  of  the 
same  powers  of  the  variable  are,  by  Art.  441,  equal.  The  equa- 
tions thus  formed  will  furnish  the  values  of  the  assumed 
numerators. 

EXAMPLES  CXIV 

Decompose  the  following  into  partial  fractions : 

^    3s*  +  7a;-4 

Solution.     As  the  divisors  of  the  denominator  are  x,  x  —  1,  and  x  +  2, 

we  assume 

3x2  +  7x-  4  =  A         B      ,      G 


x*  -\-  x2-2x       x      x-l      x  +  2 
Clearing  of  fractions  and  collecting  terms, 

3x'i  +  1x-4=(A+  B  +  C)x'2+(A  +  2B-  C)x-2A, 

whence  (Art.  441) 

A+  J5+  0  =  3, 

A  +  2B-C=1, 

-2A  =  -4. 
From  these  equations  we  have 

^4  =  2,  B  =  2,  and  C  =  -  1. 

Substituting  these  values  in  the  assumed  partial  fractions,  we  have 

3  x2  +  7  x  -  4  _  2         2 1 

x*  +  x2  -2x      x     x-l      x  +  2 


280  HIGHER  ALGEBEA 

3^-7^  +  6 
(x  -  l)3 

Solution.     As  the  divisors  of  the  denominator  are    (x  —  l)3,  (x  —  l)2, 
and  x  —  1,  we  assume 

3x2-7x+6_       A  B  C 


(x-l)3  (x-l)3      (x-l)2     x-l 

Clearing  of  fractions  and  collecting  terms, 

3  x2  -  7  x  +  6  =  Cx2  +  (B  -  2  C)x  +  .4  -  B  +  C, 

whence  (Art.  441) 

0  =  3, 

5-2C  =  -7, 

A-B+  0=6. 

From  these  equations  we  have 

C  =  S,  B--1,  xbA  A  =  2. 

Substituting  these  values  in  the  assumed  partial  fractions,  we  have 

3  **  -  7  fc  +  6  _        2  1        +3 


3. 


5. 


(X  -  l)3  (X  -  l)3      (x  -  l)2  '  x-l 

x2-2  x  +  1 

x  —  x*  x2  —  2x 

x  -h  "3  -  » 4- 1 


a*  -  a?  -  2  a2  -  7  a?  +  12 

7    ^  -  11  g  4-  26  2.x -13 

(a?-3)8      '  '   aj»  +  10aj  +  25° 

3  a:2 -4  5a?2-4o; 

'   (^  +  1)3'  "   (5x-2f 

13  a; +  10  12    £ 

'  6x3-13x2-5x'  '  fl^  +  6^  +  110+6 

.^4-3^-8  3a?-lla?  +  13a?-4 

'     x(^  +  2)2  '  '         (x>-x)(x-2f 

Case  II 

448.  When  the  denominator  is  resolvable  into  equal  or  unequal 
quadratic  factors. 

Rule.  Assume  the  given  fraction  equal  to  the  sum  of  several 
fractions  whose  numerators  have  the  form  Mx  +  N,  and  whose  de- 


DECOMPOSITION  OF  FRACTIONS  281 

nominators  are  all  the  quadratic  divisors  of  the  given  denominator, 
and  proceed  as  in  Case  I. 

Dem.  The  numerator  of  a  partial  fraction  may  contain  a  term 
with  the  first  power  of  the  variable  as  well  as  an  absolute  term, 
since  such  a  fraction,  being  in  the  form 

Mx  +  N  Mx  +  N 

or 


x2  +  ax  +  b        (x*  +  ax  +  b)n' 

in  which,  by  hypothesis,  x2  +  ax  +  b  can  not  be  factored,  is  not 
capable  of  farther  reduction. 

Moreover,  if  x  were  not  introduced  into  the  numerator,  on  clear- 
ing of  fractions  the  degree  of  the  second  member  would  be  lower 
by  two  than  the  denominator  of  the  given  fraction,  while  the  first 
member  (the  numerator  of  the  given  fraction)  may  be  lower  by 
only  one.  Then  in  equating  coefficients,  the  coefficient  of  the 
highest  power  of  x  in  the  first  member,  a  given  constant,  would  be 
placed  equal  to  0,  which  is  absurd.  It  is  for  a  similar  reason  that 
a  constant  term  must  be  introduced  into  the  numerator  of  each 
assumed  fraction. 

EXAMPLES  CXV      . 

Decompose  the  following  into  partial  fractions : 
6^  +  53  +  4 

Solution.  As  the  divisors  of  the  denominator  are  x  and  x2  +  2  x  +  2,  we 
assume 

6x2  +  5x  +  4  _A         Bx  +  C 

x3  +  2  x2  +  2x      x      x2  +  2x  +  2* 

Clearing  of  fractions  and  collecting  terms, 

6x2  +  5x  +  4  =  (>H-£)x2  +  (2A  +  C)x+2A, 

whence  (Art.  441) 

A  +  B  =  6, 

2A+C=o, 

2A  =  4. 

From  these  equations  we  have  .4  =  2,  B  —  4,  and  C  =  1. 
Substituting  these  values  in  the  assumed  partial  fractions,  we  have 

6x2  +  5x  +  4  __2  4x-+  1 


x3  +  2  x2  +  2  x      x      x2  +  2  x  +  2 


282  HIGHER  ALGEBRA 

ar3  4-  x  —  1 


2. 


(»*  +  2) 


Solution.     As  the  divisors  of  the  denominator  are  (x2  +  2)2  and  x2  +  2, 
we  assume 

x3  +  a;  -  1  _  ^4x+  ff       Cx  -f  D 

(x2  +  2)2       (x2  +  2)2       x2  +  2  ' 

Clearing  of  fractions  and  collecting  terms, 
x3  +  x  -  1  =  Cx3  +  Dx2  +  (A+  2  C)x  +  B  +  2  D,  whence  (Art.  441), 

C=l, 
I>  =  0, 
.4  +  2  C=l, 
B  +  2D=-l. 

From  these  equations  we  have  A  =  -l,  B  =  —  1,   0=1,  Z>  =  0. 
Substituting  these  values  in  the  assumed  partial  fractions,  we  have 

x3  +  x  —  1  _        x  +  1 x 

(x2  +  2)2  ~       (x2  +  2)2      x2  +  2* 

4    x*-2x  +  Z 


42  - 19  a 

X3  - 

-4ar*  +  a;  — 
1 

4 

ar5- 

_a^  +  2a;- 

2 

12 

S»  _  x  +  10 

6. 


(a,-2  + 1)2 

ar5-! 


(£C2_2#_|_l)(V+a;_|_4) 


a3  +  2  q?  +  2 
a^-1  (a;2  +  2)  (a?  +  x  +  2) 

SECTION  II  — TAYLOR'S  FORMULA 

449.  Taylor's  Formula  is  a  formula  for  developing  a  function 
of  the  sum  of  two  variables  in  terms  of  the  ascending  powers  of 
one  of  the  variables. 

450.  Factorial  n  is  the  product  of  all  the  integral  numbers 
from  1  to  n  inclusive,  and  is  written  \n.*     Thus, 

[3  =  2-3,  [5  =  2.3.4.5,  [w  =  2.  3.4...  n. 

451.  Prob.     To  produce  Taylor's  Formula. 
Solution.     Assume 

u  =f(x  +  y)  =  A  +  By+Cy2  +  Dtf  +  Ey4  -f  Ftf  -f  etc., 

*  The  notation  n !  is  also  employed. 


TAYLOR'S  FOJRMULA  283 

in  which  A,  B,  C,  etc.,  though  independent  of  y,  are  dependent  on 
x,  since  x  appears  in  the  development  only  as  involved  in  these 
coefficients.     These  coefficients  are,  therefore,  variable. 

Obtaining  the  successive  partial  derivatives  with  reference  to  y 
and  equating  them,  according  to  Art.  430,  with  the  corresponding 
successive  partial  derivatives  with  reference  to  x,  we  have 

^  =  B  +  2  Cy  +  3Dtf  +  4^  +  5Ftf  +  etc.  =^, 
dy  ax 

^  =  2  C+  2  •  3  Dy  +  3  .  4 Ey*  +  4  .  5 jy  +  etc.  =  || 

^  =  2  •  3Z)  +  2  •  3  •  AEy  +  3  .  4  .  5*y  +  etc.  =  ^, 

—  =  2  •  3  •  4  E  +  2  •  3  •  4  •  5  Fv  +  etc.  =  — [, 
dy*  dxA 

etc.,  etc.,  etc. 

Since  A,  B,  C,  etc.,  are  independent  of  y,  if  we  obtain  their 
values  for  one  value  of  y,  we  shall  have  them  (in  form)  for  all  val- 
ues of  y.     Representing  by  u'  the  value  of  u  when  y  =  0,  we  have 

a        i    t>     du1    n     d2a'l    n     d hi*  1     j-,      dVl     ,n  . 

A=w'  B=T*'  c=^2'  d=mw  E=^\£etc- 

Substituting  these  values  in  the  assumed  series,  we  have 

*/  \         i  ,  flu'     ,  d?u'y2  .  dPu'if  .  d*u'yA  .     ,n 

u  =/(*  +  j,)  =  »<  +  _,  +  _|  +  _|  +  _£  +  etc., 

which  is  Taylor's  Formula. 

452.  Sen.  As  successive  derivatives  are  represented  by  /',  /", 
/'",  etc.  (Arts.  425,  426,  and  427),  and  as  f(x  +  y)  becomes  f(x) 
when  y  =  0,  Taylor's  Formula  may  be  written 

u  =f(x  +  y)  =f(x)  +  f'(x)y  +  /"(x)|-  +f"(x)£+r(x)^  +  etc. 

This  formula  may  be  stated  as  a  theorem  as  follows : 
Taylor's  Formula  develops  u  —  f(x  -f  y)  into  a  series  in  which  the 
1st  term  is  ivhat  the  function  becomes  when  y  =  0;  the  '2d  term  is 


284  HIGHER  ALGEBRA 

y  times  ivhat  the  1st  derivative  becomes  when  y  =  0 ;  the  3d  term  is 

V2 

t^  times  what  the  second  derivative  becomes  when  y  =  0 ;  the  4:th  term 

is  ~  times  whoX  the  3d  derivative  becomes  when  y  =  0,  and  so  on. 
[3  9        ' 

EXAMPLES   CXVI 

Expand  by  Taylor's  Formula  the  following : 

i.  (x  +  yy. 

Solution,     u  =  (x  +  */)5,  u'  =  x5,   —  =  6  x4,  —  =  4  .  5  x3, 
K--*w"  dx  dx* 

$2-  =  3  •  4  •  5  x2,  ~  =  2  •  3  •  4  •  5  3,  —  =2.3.4.5. 
<Zx3  dx4  <7x° 

Substituting  these  values  in  Taylor's  Formula,  we  have 
u  =  (x  4-  2/)5  =  a,*5  -4-  5  x4y  +  10  a?3?/2  4- 10  afy3  +  5  #?/4  -4-  ^z5. 

2.    (a-?/)7.  3.    (x  +  y)k  4.    (a?  -  y)~\ 

5.    (a;  -4-  y)  *.  6.    Vx  +  y.  7.    V(a>*  -f  ?/)3. 

453.  Taylor's  Formula  is  much  used  for  developing  a  function 
of  a  single  variable  after  the  variable  has  taken  an  increment. 

EXAMPLES   CXVII 

Expand  by  Taylor's  Formula  the  following,  after  the  variable 
has  taken  an  increment  h. 

1.  2x3-x2  +  5x-ll. 

Solution.     Using  the  notation  of  Art.  452, 

/(*  +  h)  =  2(*  +  hf  -  (x  +  ny  +  5  (x  +  70  -  11, 

/(x)  =  2  xs  -  x2  +  5  x  -  11,  f  (x)  =  6  x?  -  2  x  +  5, 
/"  (x)  =  12  x  -  2,  /'"  (x)  =  12. 
Substituting  these  values  in  Taylor's  Formula,  we  have 

2  x*  -  x2  +  5  x  -  11  +  (6  x2  -  2  x  +  5)  /*  +  (6  x  -  1)  h*  +  2  7i3. 

2.  3  a5  -  2  x-2.  3.   2  x4  -  4  x3  +  a2  -  5. 

454.  In  producing  Taylor's  Formula  we  made  use  of  the  prin- 
ciple that  in  a  function  of  the  sum  of  two  variables  the  correspond- 
ing partial  derivatives  are  equal.     The  formula  will  not,  therefore, 


THE  BINOMIAL   FORMULA  285 

directly  expand  such  a  function  as  (3  x2  -f  4  y*)5,  in  which  the  cor- 
responding partial  derivatives,  on  account  of  the  coefficients  and 
exponents  of  x  and  y,  are  not  equal.  Indirectly,  however,  by 
substituting  z  for  3  x2  and  v  for  4  y3,  the  formula  will  apply ;  for 
in  that  case  the  coefficients  and  exponents  of  x  and  y  will  not 
enter  the  differentiation.  Then,  after  expansion,  the  values  of  z 
and  v  may  be  restored. 

EXAMPLES  CXVin 
Develop  by  Taylor's  Formula  the  following : 

l.   (rf  +  dy)*. 

Solution.     Substituting  z  for  x2  and  v  for  3y,  and  developing  by  Taylor's 
Formula,  we  have 

(*  +  «)*  = 

Restoring  values 


(s  +  v)$  =  z^  +  £ 2  h-±z  t*  +  TV ^  V*  -  Tf* ^  V  +  etc. 


yy  2x     8  a*      16  a;*      128x7 

2.   (2^-f-^)4-  3-    (x~*  +  2y)-\  4.    (a;-2  2/2)'. 

455.  All  of  the  above  functions,  which  are  algebraic,  can  be 
readily  expanded  without  the  use  of  Taylor's  Formula.  The 
formula  has  its  most  important  application  in  expanding  func- 
tions that  are  not  algebraic. 

SECTIOiV  III  — THE  BINOMIAL  FORMULA 

456.  The  Binomial  Formula  and  applications  with  positive 
integral  exponents  have  already  been  given  in  Arts.  155  and  159. 
The  proof  and  applications  with  any  exponent  have  been  reserved 
for  this  part  of  the  work. 

457.  Prob.     To  produce  the  Binomial  Formula. 

Solution.  Applying  Taylor's  Formula  (Art.  451)  to  the  ex- 
pansion of  the  form  (x  -\-  y)m,  we  have 

f(x  +  y)  =  (x  +  y)'»,    f(x)  =  xm,     ' 

f(x)  =  mx™1,    f'(x)  =  m(m  —  t)  xm~\ 
f'"(x)  =  m  (m  - 1 )  (m  -  2)  xm-% 
f\x)  =  m(m-  1)  (m  -  2)  (m  -  3)  xm~\  etc. 


286  HIGHER  ALGEBRA 

Substituting  these  values,  we  have 

/          x™       »  ,        .  i     ,  m(m  —  1)   m  9  „      m(m  —  l)(m  —  2)       a  _ 
(a?  +  2/)m=  scTO-f-  mxm~xy  -\ ^ J-xm~hf  -\ ± ^ ;-x1l~hf 

,         *  m  (m  -  1)  (m  -  2)  •  •  •  (m  -  n  +  2)         +1 
+  -+—  [n_1  *         2/      +  -, 

which  is  the  Binomial  Formula. 

458.  The  Binomial  Theorem.  In  the  expansion  of  a  binomial 
affected  with  any  exponent,  the  exponent  of  the  leading  letter  begins  in 
the  first  term  with  the  exponent  of  the  binomial,  and  in  each  succeeding 
term  decreases  by  1 ;  while  the  exponent  of  the  second  letter  begins  in 
the  second  term  with  1  and  in  each  succeeding  term  increases  by  1. 

The  coefficient  of  the  first  term  is  1 ;  that  of  the  second  term  is 
the  exponent  of  the  binomial ;  and  if  the  coefficient  of  any  term  be 
multiplied  by  the  exponent  of  the  leading  letter  in  that  term  arid 
divided  by  the  exponent  of  the  second  letter  increased  by  1,  the  result 
will  be  the  coefficient  of  the  next  term. 

This  is  deduced  by  inspection  from  the  Binomial  Formula. 

EXAMPLES   CXIX 
Expand  by  the  binomial  theorem  the  following : 

1.  (x2-2y)%. 

Solution.  The  theorem  (458)  applies,  not  to  the  exponents  and  coeffi- 
cients of  x  and  y  in  such  a  function  as  this,  but  to  x2  and  —  2  y  treated 
as  wholes.  To  avoid  carrying  over  into  following  terms  factors  and  signs 
that  do  not  belong  there,  it  is  best  to  inclose  the  separate  factors  in  paren- 
theses, using  the  plus  sign  between  all  the  terms,  and  reducing  afterward. 
Thus, 

(x2_2y)f=(^)t  +  (2)(a.2)4(_2y)-f(-i)(a:2)-^(_22/)2 

+  (A)(*T^(-2  2/)3+  etc., 
=  x%  -  |  x~hj  -  f  aT^/2  -  ff  x~Tys  -  etc. 

2.  (a -bf.  3.    (x-y)7.  4.   (l+x)\ 
5.    (1-2/)5.                   6.   (z-r-2/)-2.  7.   (x-y)~s. 


THE  BINOMIAL  FORMULA  287 

8.    {a-x)~\  9.  ,  or  (*  +  */)-4. 

10.    — L_.  ii.    (l_a2)l  12.    (2  + a3)*. 

13.   fx-r-X-  14    0«~2-y*)~*  15.   (V^3  +  4s!/a)4. 

1 

(1  -  a2)3 

19.   (3a-^)-3.  20.   (a2-62)l  21.  a 

V&2  —  c2^2 

22.    (1+Va)4  +  (1-V^)4-  23.    (1+V5)5  +  (1-V5)5. 

24.    (1  +  X  —  a2)4. 

Sug.     Write  in  the  form  [(1  -f  x)  —  a:2]4,  and  regard  1  +  x  as  one  term  of 
a  binomial,  and  —  x'2  as  the  other. 

25.    (a  +  b  +  c)5.  26.    (1  -  x  +  x2  -  ar5)4. 


16.    Va  +  2.  17.    -    -L_98»  18.    (a2 -a;2 


CHAPTER   XXI 
LOGARITHMS 

459.  The  Logarithm  of  a  Number  is  the  exponent  by  which  a 
fixed  number  is  affected  to  produce  any  required  number.  The 
fixed  number  is  called  the  Base. 

Thus,  let  4  be  the  base  ;  then 

3  is  the  logarithm  of  64  to  base  4,  since  43  =  G4 
2  is  the  logarithm  of  16  to  base  4,  since  42  =  16 
1  is  the  logarithm  of  4  to  base  4,  since  41  =  4 
0  is  the  logarithm  of    1  to  base  4,  since  4°    =    1 

—  1  is  the  logarithm  of    \  to  base  4,  since  4_1  =    J 

—  2  is  the  logarithm  of  j1^  to  base  4,  since  4-2  =  TV 

460.  Logarithms  are  in  one  system  or  another  according  to  the 
base  assumed.  Only  two  systems  are  in  common  use,  viz.,  the 
system  whose  base  is  10,  called  the  Briggean,  Briggs,  or  Common 
System,  and  the  system  whose  base  is  2.71828 +,  called  the 
Napierian,  Natural,  or  Hyperbolic  System.  The  Common  System 
is  used  to  facilitate  numerical  calculations,  and  the  Napierian 
System  is  much  used  in  abstract  mathematical  discussions. 

When  necessary  to  distinguish  different  systems,  the  bases  are  written  as 
subscripts  to  the  abbreviation  log.  Thus,  k  —  loga  n  signifies  that  k  is  the 
logarithm  of  n  in  the  system  whose  base  is  a.  The  student  should  clearly 
understand  that,  by  definition,  k  =  loga  n  expresses  the  same  relation  between 
k  and  n  as  does  ak  =  n. 

461.  Theorem.     The  logarithm  of 1  is  0  in  all  systems. 
For,  by  Art.  70,  a0  =  1  for  all  values  of  a. 

462.  Theorem.  In  any  system  whose  base  is  greater  than  1  the 
logarithm  ofOis—  oo. 

For,  a  being  the  base,  a-00  =  —  =  0.     Therefore  log  0  =  —  oo. 

a? 

288 


LOGARITHMS  289 

463.  Theorem.  Neither  1  nor  any  negative  number  can  be  used 
us  tlw  fxisc  of  a  system  of  logarithms. 

For  with  1  as  a  base  we  can  represent  no  other  number  than  1 
by  its  exponents,  1  with  any  exponent  being  1. 

Again,  with  a  negative  base,  odd  exponents  would  give  only- 
negative  numbers,  and  the  corresponding  positive  numbers  would 
have  no  logarithms.  For  example,  with  —  2  as  a  base,  3  would 
be  the  logarithm  of  —  8,  since  (—  2)3  =  —  8 ;  but  +  8  would  have 
no  logarithm,  since  it  cannot  be  produced  by  affecting  —  2  with 
any  exponent. 

464.  Theorem.     Negative  numbers,  as  such,  have  no  logarithms. 

For  a  negative  number  cannot  be  produced  by  affecting  a 
positive  base  with  any  exponent. 

465.  Sen.  When  negative  numbers  occur  in  computation  by 
logarithms,  they  are  treated  as  if  they  were  positive,  and  the  sign 
of  the  result  is  determined  by  the  number  of  negative  factors. 
It  is  usual  to  write  a  subscript  n  after  the  logarithm  of  a  negative 
number.  Thus,  1.425673n  indicates  that  the  number  of  which 
1.4U5G73  is  the  logarithm  is,  in  itself,  negative. 

466.  The  most  important  use  of  logarithms  is  to  facilitate  the 
multiplication,  division,  involution,  and  evolution  of  numbers 
containing  several  figures.  The  processes  depend  on  the  follow- 
ing principles. 

467.  Theorem.  The  logarithm  of  the  product  of  two  numbers  is 
the  sum  of  their  logarithms. 

Dem.  Let  a  be  the  base  of  the  system,  and  p  and  q  any  two 
numbers  whose  logarithms  are  x  and  y  respectively.  Then,  by 
definition, 

p  =  ax  and  q  =  ay. 

Multiplying  together  the  corresponding  members  of  these  equa- 
tions, we  have 


pq  —  ax  x  ay  =  a 


*+v 


that  is,  log  jx/  =  *  +  y  =  logjr?  +  log  q. 

downey's  alg.  — 19 


290  HIGHER  ALGEBRA 

468.  Cor.     In  the  same  way  it  may  be  shown  that 

logpgr  •••  =  logp  -f  logq  +  logr  -f  .... 

469.  Theorem.     The  logarithm  of  the  quotient  of  two  numbers 
is  the  logarithm  of  the  dividend  minus  the  logarithm  of  the  divisor. 

Dem.     Let  a  be  the  base  of  the  system,  and  p  and  q  any  two 
numbers  whose  logarithms  are  x  and  y  respectively.     Then,  by 


definition, 

Dividing, 

we 

have 

P  = 
P. 

--ax 

a* 
~~a? 

and  q  = 
=  ax~v ; 

--  ay. 

that  is, 

lOg^ 

Q 

=  X- 

-y- 

=  logp  - 

-lo 

470.  Theorem.  The  logarithm  of  a  power  of  a  number  is  the 
logarithm  of  the  number  multiplied  by  the  index  of  the  power. 

Dem.     Let  a  be  the  base,  and  x  the  logarithm  of  p. 
Then,  by  definition,  p  =  a*. 

liaising  both  members  to  any  power,  as  the  nth,  we  have 
pn  =  (ax)n  =  anx ; 
that  is,  log pn  =  nx  =  n  logp. 

471.  Theorem.  The  logarithm  of  any  root  of  a  number  is  the 
logarithm  of  the  number  divided  by  the  index  of  the  root. 

Dem.     Let  a  be  the  base,  and  x  the  logarithm  of  p. 

Then,  by  definition,  p  =  ax. 

Extracting  the  nth  root  of  both  members,  we  have 

X 

■\/p  as  -\/ax  =  an ; 

that  iS,  ]0g-w;=-=^- 

n         n 

472.  It  is  thus  seen  that  by  the  use  of  logarithms  the  opera- 
tions of  multiplication  are  replaced  by  those  of  addition,  division 
by  subtraction,  involution  by  a  single  multiplication,  and  evolu- 
tion by  a  single  division. 


LOGARITHMS  291 

473.  Only  those  numbers  that  are  exact  powers  of  the  base 
have  integral  logarithms;  hence  the  logarithms  of  most  numbers 
consist  of  two  parts,  an  integer  and  a  decimal  fraction.  It  is 
found  convenient  to  write  a  logarithm  so  that  its  fractional  part 
shall  be  positive,  and  its  integral  part  positive  or  negative,  as  the 
case  may  be. 

474.  The  Characteristic  of  a  Logarithm  is  its  integral  part,  and 
the  Mantissa  of  a  Logarithm  its  fractional  part,  when  the  logarithm 
is  so  written  that  the  fractional  part  is  positive. 

475.  Prob.  To  ascertain  the  law  of  the  characteristics  of  loga- 
rithms in  the  common  system. 

Solution.         Since  10°  =  1,        log        1=0; 

since  101  =  10,      log      10=1; 

since  102  =  100,    log    100  =  2 ; 

since  103  =  1000,  log  1000  =  3 ; 


Also, 


since  10"3  =  -i-  =  .001,  log  .001  =  -  3 ; 
10'5 

etc.,  etc.,  etc. 

Thus  it  is  seen,  and  this  is  true  in  any  system,  that  when  the 
logarithms  are  in  arithmetical  progression,  the  numbers  are  in 
geometrical  progression. 

Now  the  logarithm  of  any  number  between  1  and  10  is  between 
0  and  1,  i.e.,  0  +  a  fraction ;  the  logarithm  of  any  number  be- 
tween 10  and  100  is  between  1  and  2,  i.e.,  1  +  a  fraction ;  the 
logarithm  of  any  number  between  100  and  1000  is  between  2  and 
3,  i.e.,  2-f  a  fraction ;  and  so  on.  Hence  the  characteristic  of  the 
logarithm  of  a  number  >  1  is  one  less  than  the  number  of  figures  in 
the  integral  part  of  the  number. 

Again,  the  logarithm  of  any  number  between  1  and  .1  is  be- 
tween  0  and  —  1,    i.e.,  —  1  +  a  fraction ;  the  logarithm  of  any 


etc.,         etc., 

etc. 

since  10"1  =  —  =  .1, 
10        ' 

log      .1=-1; 

since  10"2  =  — =  .01, 
102 

log    .01  =  -2; 

292  HIGHER  ALGEBRA 

number  between  .1  and  .01  is  between  —  1  and  —  2,  i.e.,  —  2  +  a 
fraction ;  the  logarithm  of  any  number  between  .01  and  .001  is 
between  —  2  and  —  3,  i.e.,  —  3  -f  a  fraction  ;  and  so  on.  Hence 
the  characteristic  of  a  logarithm  of  a  decimal  fraction  is  negative, 
and  numerically  1  greater  them  the  number  of  0's  before  the  first 
significant  figure. 

476.  Characteristics  are  not  written  in  tables  of  logarithms, 
but  are  to  be  supplied  according  to  the  above  principles. 

477.  When  the  characteristic  of  a  logarithm  is  negative,  the 
minus  sign  is  written,  not  before  it,  but  over  it,  since  the  charac- 
teristic alone  is  negative.     Thus  log  .00167  =  3.222716. 

478.  Theorem.  Whatever  change  be  made  in  the  position  of  the 
decimal  point  in  a  number,  the  mantissa  of  its  logarithm  in  the 
common  system  remains  the  same. 

Dem.  Changing  the  position  of  the  decimal  point  is  equiva- 
lent to  multiplying  or  dividing  by  some  power  of  10.     Now 

log  (k  x  10B)  rs log* -f  log  10n  =  log  k  +  n  log  10  =  log  k  +  n, 
log  (k  -i-  10n)  =  log  k  —  log  10*  =  log  k  —  n  log  10  =  log  k  —  n. 

Hence  the  logarithms  of  k,  k  x  10n,  and  k  -=-  10n  differ  only  by 
an  integer  n\  i.e.,  the  mantissa?  of  the  logarithms  of  these  num- 
bers, which  differ  only  in  the  position  of  the  decimal  point,  are 
the  same. 

479.  Sch.  The  characteristic,  in  the  common  system,  charac- 
terizes the  number  in  showing  between  what  two  consecutive 
powers  of  10  it  lies,  and  depends,  therefore,  simply  on  the  posi- 
tion of  the  decimal  point;  while  the  mantissa,  which  means  an 
addition,  is  the  part  added  to  the  characteristic  to  give  the 
approximate  logarithm,  and  depends  simply  on  the  sequence  of 
figures. 

480.  Theorem.  The  differential  of  the  logarithm  of  a  variable  is 
the  differential  of  the  variable  multiplied  by  a  constant,  called  the 
modulus  of  the  system,  divided  by  the  variable. 


LOGARITHMS  293 

In  the  Napierian  system,  the  modulus  being  1,  the  differential  of 
the  logarithm  of  a  variable  is  the  differential  of  the  variable  divided 
by  the  variable. 

Dem.     Let  y  =  xn,  (1) 

n  being  any  arbitrary  constant.     Then  (Art.  420), 

dy  =  nxn  1  dx  =  n  -  dx, 
u  x     ' 

dy        dx 

From  (1)  we  have,  by  Art.  470, 

log  y  =  n  log  x. 
Hence  from  Art.  413,  whatever  the  differentials  of  logy  and 

d(logy)=nd(\ogx).  (3) 

Dividing  (3)  by  (2)  to  eliminate  n, 

dQogy)  =d(\ogx) 
dy  dx 

V  % 

Hence  the  ratio  of  d(\ogy)  to_^is  the  same  as  that  of  d(\ogx) 

dx 

to  — ,  whatever  the  values  of  x  and  y. 

Represent  this  constant  ratio  by  m.     Then 

d(logy)=1^1,  andd(logz)  =  ^. 
y  x 

481.  The  Logarithmic  Series  is  the  development  of  log(l  +  a?) 
in  ascending  powers  of  x.  It  reveals  several  important  properties 
of  logarithms,  and,  when  rendered  convergent,  furnishes  a  means 
of  computing  the  logarithms  of  numbers. 

482.  Prob.    To  produce  the  logarithmic  series. 
Solution.     Let  u  =  log  (z  -f  x)  ;  then  (Art.  451) 


u' 

=  logz 

du' 
'    dz~ 

m 

'■     > 
z 

dV_ 

dz1 

m 

dz* 

2  m 

d*u'_ 
dz> 

2 

•3m 

z*    ' 

etc. 

294  HIGHER  ALGEBRA 

Substituting  these  values  in  Taylor's  Formula  (Art.  451),  we  have 

«  =  log(*  +  »)  =  log*  +  ™x-™,f +  ££-»!  £  +  eto. 

z         z2  2      z3  3      z4  4 

Making  z  =  1,  this  becomes 

log  (1  +  x)  =  m(x  - 1  + 1  - 14  +  etc.), 
which  is  the  logarithmic  series. 

483.  Sch.  1.  In  the  above  series,  if  we  let  x  remain  the  same 
and  express  the  logarithms  of  1  +  x  in  two  different  systems 
whose  bases  are  a  and  b,  loga  (1  -f  x)  is  not  equal  to  log6  (1  +  x). 
But  the  series  in  the  parenthesis  is  the  same  in  the  two  cases ; 
hence  m  is  different  in  the  two  cases,  i.e.,  m  depends  on  the  base 
and  characterizes  the  system.  This  factor,  which  is  constant  in 
the  same  system  but  different  in  different  systems,  has  been 
named  the  modulus. 

484.  Sch.  2.  It  is  evident  that,  in  establishing  a  system  of 
logarithms,  we  may  either  select  the  modulus  and  let  the  base  be 
determined  by  the  mutual  relation,  or  select  the  base  and  let  the 
modulus  be  determined  by  the  mutual  relation.  In  the  Napierian 
system  (so  called)  1  was  selected  as  the  modulus,  and  this  gave 
for  the  base  2.718281828 -f-  ;  while  in  the  Briggs  system  10  was 
selected  as  the  base,  and  this  gave  for  the  modulus  .43429448  +  . 

485.  Theorem.  TJie  logarithms  of  the  same  number  in  different 
systems  are  to  each  other  as  the  moduli  of  those  systems. 

Dem.  Letting  a  and  a!  be  the  bases,  and  m  and  m'  the  moduli 
of  two  systems,  we  have,  from  Art.  482, 

log0(l  +  z)  =  m(z-^  +  J-J  +  etc.),  (1) 

log„.(l  +  *)  =  m\x  - 1  + 1  -  J  +  etc.).  (2) 


Dividing  (1)  by  (2), 

log.(l  +  x)  _  m 
log,  (1  +  x)      to'- 


(3) 


LOGARITHMS  295 

486.  Cor.  The  logarithm  of  a  number  in  the  Napierian  system 
multiplied  by  the  modulus  of  any  other  system  will  give  the  logarithm 
of  the  same  number  in  the  latter  system. 

487.  Prob.  To  render  the  Napierian  logarithmic  series  con- 
vergent. 

Solution.  As  the  logarithmic  series  (Art.  482)  is  divergent 
for  x  >  1,  it  cannot,  nntil  transformed  into  a  convergent  series,  be 
used  for  computing  the  logarithms  of  numbers. 

For  the  Napierian  system,  whose  base  is  usually  represented 
by  e,  the  series  becomes,  when  x  is  positive, 

/>»2  /y&  /v»4  a») 

log.(l+*)  =  *-f  +  |-f  +  |-etc,  (1) 

and  when  x  is  negative, 

/>*2  /smO  /-}%A  /yj> 

log.(l-x)=-«-|-J-|-|-ete.  (2) 

Subtracting  (2)  from  (1),  we  have 
log.(l  +  x)  -  log.(l  -  x)  =  loge  t±|  =  2(W|  +  !+  etc.Y 

Letting  ^ = — , 

1  —  x      q 

which  gives  by  solving  the  equation 

-    p-q 


and  substituting, 


P  +  q' 


l0gef  =logep-\ogeq 


or 


which  is  seen  to  be  a  convergent  series. 

488.    Prob.     To  compute  the  logarithms  of  1,  2,  3,  4,  5,  etc. 

Solution.  It  is  necessary  to  compute  the  logarithms  of  prime 
numbers  only,  since  the  logarithm  of  a  composite  number  is  equal 
to  the  sum  of  the  logarithms  of  its  factors  (Art.  467). 


296  HIGHER  ALGEBRA 

The  logarithm  of  1  is  0  (Art.  461). 

To  find  the  Napierian  log  2,  in  (A)  make  p  =  2  and  q  =  1. 
To  find  the  Napierian  log  3,  make  p  =  3  and  q  =  2,  and  so  on. 
We  thus  have 

log^log.^g+^  +  ^  +  ^  +  ^+otc.) 

0.69314718, 

log.3  =  loge2  +  2(1  +  ^  +  ^  +  -^+  oto.)  =  1.09861228, 

log.  4  =  2  log.  2  =  1.38629436, 

108,6  =  108,4  +  2^  +  ^  +  ^  +  ^  +  610.)=  1.60943790, 

and  so  on. 

The  common  logarithms  of  the  same  numbers  may  now  be 
found  by  multiplying  the  Napierian  logarithms  by  the  modulus 
of  the  common  system  (Art.  486), 

489.  Tables  of  such  logarithms  have  been  prepared  with  great 
care  and  are  easily  obtainable  by  the  student  and  the  practical 
computer.  Bremiker's  for  six  places  and  Vega's  for  seven  are 
standard  tables.  For  some  refined  astronomical  calculations  ten- 
place  logarithms  are  necessary,  while  for  many  purposes  four- 
place  and  five-place  logarithms  are  sufficient. 

490.  Theorem.  The  modulus  of  any  system  is  the  reciprocal  of 
the  Napierian  logarithm  of  the  base  of  that  system,  and  in  the 
common  system  is  .43429448 ;  and  the  Napierian  base  is  the  number 
of  which  the  modulus  of  any  system  is  the  logarithm  in  that  system, 
and  is  2.71821828. 

Dem.  In  equation  (3)  of  Art.  485,  letting  1  +  x  =  a  =  base, 
we  have 

log_  a      m  1 

— ^-  =  -,  or  m  =  - 

logea      1  logea 

In  the  common  system  this  becomes 

m  =  — —  = = =  .43429448. 

log,  10      loge2  +  loge5     2.30258508 


LOGARITHMS  297 

Again,  by  the  same  equation, 

log10e^7ft 
logee      1 
or  log10  e  =  m  loge  e  =  m  =  .43429448. 

Finding  from  a  table  of  common  logarithms  the  number  of 
which  .43429448  is  the  logarithm,  we  have 
e  =  2.71821828. 

491.  An   Exponential   Equation   is   an   equation  in  which   the 
unknown  quantity  occurs  as  an  exponent. 

492.  Prob.     To  solve  an  exponential  equation. 
Solution.     By  definition  the  equation  has  the  form 

ax  =  b. 
Taking  the  logarithms  of  both  members,  we  have 
x  log  a  =  log  6, 

whence  x  =  r-^— 

log  a 

493.  Prob.     The  principal  p,  rate  r,  and  time  t  being  given,  to 
find  the  amount  a,  at  compound  interest. 

Solution.     We  have 

for  1  year,    a  as  p  -f  pr  =  p  (1  +  r), 
for  2  years,  a  =  p  (1  +  r)  +  p  (1  +  r)  r  =  p  (1  +  r)2, 
for  3  years,  a  =  p(l  +  r)2  +p(l  +  r)2r=p(l  +  r)*, 
and  so  on.     Hence,  for  £  years, 

a  =  p(l+ry.  (1) 

Taking  the  logarithms  of  both  members,  we  have 

log  a  =  log  p  +  t  log  (1  +  r).  (2) 

494.  Cor.   i.     If  the  interest  is  compounded  m  times  annually, 
formulas  (1)  and  (2)  become 

log  a  =  log  ?>  -f  mt  log  f  1  +  —  ]< 


mt 

i 


298  HIGHER  ALGEBRA 

Cor.  2.     From  (2)  we  deduce 

log p  =  \oga-t log  (1  +  r), 

log(l  +  r)  =  loga-log*, 

log  (1  +  r) 

495.    Prob.     To  find  the  present  worth,  S,  of  an  annuity,  a,  run^ 
nina  t  years. 

Solution.     Money  being  worth  r  per  cent, 

the  present  worth  of  1st  payment  = 


1  +  r 
the  present  worth  of  2d  payment 


(l  +  r)s 


the  present  worth  of  3d  payment  = 


(1+r)3 
and  so  on.     Hence  the  present  worth  of  the  whole  is  the  sum  of 

a  geometrical  progression   in  which  the  first  term  is  — - — ,  the 

1  1  +  r 

ratio  is ,  and  the  number  of  terms  is  t.     Substituting  in  the 

formula  of  Art.  327  and  reducing,  we  have 

q[(l  +  ry-l] 

r(i  +  ry    ' 

whence  log  S  =  log  a  +  log  [(1  +  r)1  —  1]  —  log  r  —  t  log  (1  +  r). 

EXAMPLES   CXX 
Express  the  following  operations  by  logarithms  : 


a2  —  ar2 


Solution.  y  =  J^^  =  Ka  +  x)(a-x)\\ 

*  l  +  x      \         l  +  x         I 

Hence,        log  y  =  $  [log  {a  +  x)  +  log  (a  -  x)  -  log  (1  +  x)]. 


2.    y  =  **(l-aO*.  3    y=\P 


—  a) 
be 


4.   v/=  '/a&'c4.  5.   y=    la* -a4l 


6.  y  = 


LOGARITHMS  299 

-y/x  —  x2 


'-V? 


8    y-,/^'/3-i3'  V     ^-x-2    ' 


Va  Va2  —  62 
Differentiate  the  following  : 


10.   y  =  log  VI  +  «. 


Solution,     y  =  log  Vl  +  x  =  log  (1  +  x) 2  =  |  log  (1  +  sc). 
Hence,  by  Arts.  413  and  480, 

dy         mdx 


2(1  +  x) 

Or,  differentiating  without  first  changing  the  form,  we  have  (Arts.  421 

and  480), 

mdx 


,       2  VI  +  x         mdx 
dy- 


Vl  +  x  2(1  +  x) 

11.   y  =  log  ax.  12.   y  =  log  (1  —  x). 

13.   y  =  logar\  14.   y  =  log- 
ic 

15.   ?/ =  log  (a2  -  ar>).  16.   y  =  log  (1  +  x2)2. 

Solve  the  following  equations : 

17.   ahx  =  c.  18.   ax  =  -- 

19.   bacx  =  d.  20.   102*  -  6  •  10*  =  7. 


10*+'  =  1000,  f  a*+" 

21.    J  22 

10*""  =  100. 


1  &*-»  =  n. 


496.  Following  are  two  sample  pages  from  a  table  of  logarithms. 
The  characteristics  are  not  written  in  the  table,  but  are  to  be  sup- 
plied by  the  principles  of  Art.  475. 

497.  To  find  from  the  table  the  mantissa  of  the  logarithm  of  a 
number. 

In  all  cases  the  decimal  point  is  disregarded,  as  it  has  to  do 
only  with  the  characteristic  (Art.  478).     The  logarithms  of  the 


300 


HIGHER  ALGEBRA 


Logarithms  of  Numbers 


N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

Diff. 

770 

880491 

6547 

6004 

6660 

6716 

6773 

6829 

6885 

6942 

6998 

57 

771 

7054 

7111 

7167 

7223 

7280 

7336 

7392 

7449 

7505 

7561 

1   6 

772 

7617 

7674 

7730 

7786 

7842 

7898 

7955 

8011 

80(57 

8123 

2  11 

773 

8179 

8236 

8292 

8348 

8404 

84(50 

851(5 

8573 

8629 

8(585 

3  17 

774 

8741 

8797 

8853 

8909 

8965 

9021 

9077 

9134 

9190 

924(5 

4  23 

775 

9302 

9358 

9414 

9470 

9526 

9582 

9638 

9694 

9750 

980(3 

5  29 

776 

9862 

9918 

9974 

**30 

**86 

*141 

•197 

*253 

♦309 

*365 

6  34 

777 

890421 

0477 

0533 

0589 

0(545 

0700 

075(5 

0812 

0868 

0924 

7  40 

778 

0980 

1035 

1091 

1147 

1203 

1259 

1314 

1370 

1426 

1482 

8  4(5 

779 

1537 

1593 

1649 

1705 

1760 

1816 

1872 

1928 

1983 

2039 

9  51 

780 

892095 

2150 

2206 

2262 

2317 

2373 

2429 

2484 

2540 

2595 

56 

781 

2651 

2707 

2762 

2818 

2873 

2929 

2985 

3040 

3096 

3151 

1   6 

782 

3207 

3262 

3318 

3373 

3429 

3484 

3540 

3595 

3651 

370(5 

2  11 

783 

3762 

3817 

3873 

3928 

3984 

4039 

4094 

4150 

4205 

4261 

3  17 

784 

4316 

4371 

4427 

4482 

4538 

4593 

4648 

4704 

4759 

4814 

4  22 

785 

4870 

4925 

4980 

5036 

5091 

5146 

5201 

5257 

5312 

5367 

5  28 

786 

5423 

5478 

5533 

5588 

5644 

5(599 

5754 

5809 

5864 

5920 

6  34 

787 

5975 

6030 

6085 

6140 

6195 

(5251 

630(5 

63(51 

6416 

6471 

7  39 

788 

6526 

6581 

663a 

6692 

6747 

6802 

6857 

6912 

69(57 

7022 

8  45 

789 

7077 

7132 

7187 

7242 

7297 

7352 

7407 

7462 

7517 

7572 

9  50 

790 

897627 

7682 

7737 

7792 

7847 

7902 

7957 

8012 

8067 

8122 

55 

791 

8176 

8231 

8286 

8341 

8396 

8451 

850(5 

8561 

8(515 

8670 

1   6 

792 

8725 

8780 

8835 

8890 

8944 

8999 

9054 

9109 

9164 

9218 

2  11 

793 

9273 

9328 

9383 

9137 

9492 

9547 

9602 

9656 

9711 

976(5 

3  17 

794 

9821 

9875 

9930 

9985 

**39 

**94 

*149 

*203 

*258 

*312 

4  22 

795 

900367 

0422 

0476 

0531 

0586 

0(540 

0695 

0749 

0804 

0859 

5  28 

796 

0913 

01)68 

1022 

1077 

1131 

1186 

1240 

1295 

1349 

1404 

6  33 

797 

1458 

1513 

1567 

1622 

1676 

1731 

1785 

1840 

1894 

1948 

7  39 

798 

2003 

2057 

2112 

2166 

2221 

2275 

2329 

2384 

2438 

2492 

8  44 

799 

2547 

2601 

2655 

2710 

2764 

2818 

2873 

2927 

2981 

3036 

9  50 

800 

9013090 

3144 

3199 

3253 

3307 

3361 

3416 

3470 

3524 

3578 

54 

801 

3633 

3687 

3741 

3795 

3849 

3904 

3958 

4012 

4066 

4120 

1   5 

802 

4174 

4229 

4283 

4337 

4391 

4445 

4499 

4553 

4607 

46(31 

2  11 

803 

4716 

4770 

4824 

4878 

4932 

4986 

5040 

5094 

5148 

5202 

3  16 

804 

5256 

5310 

5364 

5418 

5472 

5526 

5580 

5634 

5688 

5742 

4  22 

805 

5796 

5850 

5904 

5958 

6012 

60(56 

6119 

6173 

6227 

6281 

5  27 

806 

6335 

6389 

6143 

6497 

6551 

6604 

6658 

6712 

6766 

6820 

6  32 

807 

6874 

6927 

6981 

7035 

7089 

7143 

7195 

7250 

7304 

7358 

7  38 

808 

7411 

7465 

7519 

7573 

7626 

7680 

7734 

7787 

7841 

7895 

8  43 

809 

7949 

8002 

8056 

8110 

8163 

8217 

8270 

8324 

8378 

8431 

9  49 

810 

908485 

8539 

8592 

8646 

8699 

8753 

8807 

8860 

8914 

8967 

53 

811 

9021 

9074 

9128 

9181 

9235 

9289 

9342 

9396 

9449 

9503 

1   5  . 

812 

9556 

9610 

9663 

9716 

9770 

9823 

9877 

9930 

9984 

**37 

2  11 

813 

910091 

0144 

0197 

0251 

0304 

0358 

0411 

0464 

0518 

0571 

3  16 

814 

0624 

0678 

0731 

0784 

0838 

0891 

0944 

0998 

1051 

1104 

4  21 

815 

1158 

1211 

1264 

1317 

1371 

1424 

1477 

1530 

1584 

1637 

5  27 

816 

1690 

1743 

1797 

1850 

1903 

1956 

2009 

2063 

2116 

2169 

6  32 

817 

2222 

2275 

2328 

2381 

2435 

2488 

2541 

2594 

2(547 

2700 

7  37 

818 

2753 

2806 

2859 

2913 

296(5 

3019 

3072 

3125 

3178 

3231 

8  42 

819 

3284 

3337 

3390 

3443 

3490 

3549 

3(502 

3(555 

3708 

3761 

9  48 

N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

Diff. 

LOG  A  HIT  II  MS 


301 


Logarithms  of  Numbers 


N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

Diff. 

820 

913814 

3867 

3920 

3973 

4026 

4079 

4132 

4184 

4237 

4290 

53 

821 

4343 

4396 

4449 

4502 

4555 

4(508 

4(5(50 

4713 

47(5(5 

4819 

1   5 

822 

4S72 

4925 

41177 

5030 

5083 

5136 

5189 

5241 

6294 

5347 

2  11 

828 

5400 

6463 

5505 

5558 

6611 

5664 

5710 

5769 

5822 

5875 

3  10 

824- 

5927 

6980 

6033 

6085 

6138 

6191 

6243 

6296 

6349 

6401 

4  21 

825 

6454 

6507 

6669 

6612 

6664 

6717 

6770 

(5822 

6875 

(5927 

5  27 

826 

6960 

7033 

7085 

7138 

7190 

7243 

7295 

7348 

7400 

7453 

6  32 

827 

7606 

7668 

7611 

7663 

771(5 

7768 

7820 

7873 

7925 

7978 

7  37 

828 

8030 

8083 

8135 

8188 

8240 

8293 

8345 

8397 

8450 

8502 

8  42 

829 

8555 

8607 

8669 

8712 

8764 

8816 

8869 

8921 

8973 

902(5 

9  48 

830 

919078 

9130 

9183 

9235 

92S7 

9340 

9392 

9444 

9496 

9549 

52 

831 

9601 

9663 

9706 

975S 

9810 

9862 

9914 

9967 

**19 

**71 

1   5 

832 

920123 

0176 

0228 

0280 

0332 

0384 

0436 

0489 

0541 

0593 

2  10 

833 

0645 

0697 

0749 

0801 

0853 

0900 

0958 

1010 

1002 

1114 

3  10 

834 

1166 

1218 

1270 

1322 

1374 

1420 

1478 

1530 

1582 

1634 

4  21 

835 

1686 

1738 

1790 

1842 

1894 

1946 

1998 

2050 

2102 

2154 

5  26 

836 

2206 

2258 

2310 

2362 

2414 

2466 

2518 

2570 

2622 

2674 

6  31 

837 

2725 

2777 

2821) 

2881 

2933 

29S5 

3037 

3089 

3140 

3192 

7  36 

838 

3244 

3296 

3348 

3399 

3451 

3503 

3555 

3007 

3658 

3710 

8  42 

839 

3762 

3814 

3865 

3917 

39(59 

4021 

4072 

4124 

4176 

4228 

9  47 

840 

924279 

4331 

4383 

4434 

4486 

4538 

4589 

4041 

4693 

4744 

51 

841 

4796 

4848 

4899 

4951 

5003 

5054 

510(5 

5157 

5209 

5201 

1   5 

842 

5312 

5304 

5415 

6467 

5518 

5570 

5(521 

5073 

5725 

5776 

2  10 

843 

5828 

5879 

5931 

6982 

6034 

6085 

(5137 

0188 

0240 

6291 

3  15 

844 

6342 

6394 

6445 

(J497 

6548 

(5000 

0651 

(5702 

6754 

6805 

4  20 

845 

6867 

6908 

6969 

701.1 

7002 

7114 

7165 

7216 

7268 

7319 

5  20 

846 

7370 

7422 

7473 

7521 

757(5 

7027 

7678 

7730 

7781 

7832 

6  31 

847 

7883 

7935 

7986 

8037 

8088 

8140 

8191 

8242 

8293 

8346 

7  30 

848 

8366 

8447 

8498 

8649 

8601 

8652 

8703 

8754 

8805 

8857 

8  41 

849 

8908 

8969 

9010 

9061 

9112 

9163 

9215 

9266 

9317 

9368 

9  46 

850 

929419 

9470 

9521 

9572 

9623 

9674 

9726 

9776 

9827 

9879 

50 

851 

9930 

9981 

**32 

**83 

*134 

*185 

*236 

*287 

♦338 

*389 

1   5 

852 

9:50440 

0491 

0542 

0592 

(M543 

0694 

0745 

0790 

0847 

0898 

2  10 

853 

0949 

1000 

1051 

1102 

1153 

1204 

1254 

1305 

135(5 

1407 

3  15 

854 

1458 

1509 

1560 

1610 

1661 

1712 

1763 

1814 

18(55 

1915 

4  20 

855 

1908 

2017 

2068 

2118 

2169 

2220 

2271 

2322 

2372 

2423 

5  25 

856 

2474 

2524 

2575 

2626 

2677 

2727 

2778 

2829 

2879 

2930 

6  30 

857 

2981 

3031 

3082 

3133 

3183 

3234 

3285 

3335 

3380 

3437 

7  35 

868 

3487 

3638 

3589 

3639 

3690 

3740 

3791 

3841 

3892 

3943 

8  40 

859 

3993 

4044 

4094 

4145 

4195 

4246 

4296 

4347 

4397 

4448 

9  45 

860 

934496 

4549 

4599 

4650 

4700 

4751 

4801 

4852 

4902 

4953 

49 

861 

6008 

5054 

5104 

5154 

5205 

5255 

5300 

5350 

5400 

5457 

1   5 

862 

5507 

6668 

5008 

5658 

570! » 

5759 

6809 

58(50 

5910 

59(50 

2  10 

803 

6011 

6061 

6111 

6162 

6212 

(52(52 

6313 

6363 

0413 

(5463 

3  15 

864 

6614 

6664 

6614 

6666 

(5715 

(57(55 

6815 

6865 

(591(5 

6966 

4  20 

865 

7016 

7066 

7117 

7167 

7217 

7267 

7317 

73(57 

7418 

7468 

5  25 

866 

7.-»  IS 

7668 

7618 

7008 

7718 

7769 

7S19 

78(59 

7919 

7969 

6  29 

867 

8019 

8069 

8119 

8169 

S219 

8269 

S320 

S370 

8420 

8470 

7  34 

868 

8620 

8670 

8620 

8670 

8720 

8770 

SS20 

8870 

.Si  120 

S970 

8  39 

869 

9020 

IK  170 

9120 

9170 

9220 

9270 

9320 

9369 

9419 

9469 

9  44 

N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

Diff. 

302  HIGHER  ALGEBRA 

numbers  78,  780,  7800,  7.8,  .78,  .078,  .0078,  all  have  the  same 
mantissa. 

In  all  cases  the  first  three  figures  of  the  number  are  found  in 
the  column  headed  N,  and  the  mantissa  of  the  logarithm  is  found 
in  the  same  horizontal  line  with  these. 

If  the  number  has  but  three  figures,  the  mantissa  is  in  the 
column  headed  O.     Thus  log  793  =  2.899273. 

If  the  number  has  four  figures,  the  last  four  figures  of  the 
mantissa  are  in  the  column  headed  with  the  fourth  figure  of 
the  number,  and  the  two  initial  figures  are  in  the  column  headed 
O.  The  initial  figures  are  never  taken  from  a  lower  line  unless 
asterisks  occupy  one  or  more  places  of  the  last  four  figures  of  the 
mantissa,  in  which  case  the  places  of  the  asterisks  are  to  be  sup- 
plied with  0's.     Thus, 

log  8426  =  3.925621,  and  log  7945  =  3.900094. 

If  the  number  has  more  than  four  figures,  the  mantissa  of  the 
logarithm  for  the  first  four  figures  is  found  as  just  explained,  and 
additions  are  made  to  it  for  the  other  figures  by  means  of  one  of 
the  auxiliary  tables  in  the  margin,  the  one  to  be  used  being 
headed  with  the  difference  between  the  mantissa  just  found  and 
the  one  next  larger  in  the  table.  The  correction  for  the  5th 
figure  of  the  number  is  as  given  in  the  auxiliary  table ;  the  cor- 
rection for  the  6th  figure  is  one  tenth  of  that  given  in  the  same 
table,  and  so  on.  Thus,  to  find  the  logarithm  of  8576479,  we  pro- 
ceed as  follows  : 

Mantissa  of  log  8576  =    .933285 

Correction  for  4  in  5th  place  =  20 

Correction  for  7  in  6th  place  =  35 

Correction  for  9  in  7th  place  =  45 

Log  8576479  =  6.933309 

The  corrections  are  found  in  the  auxiliary  table  headed  50, 
this  being  the  difference  (in  the  last  two  decimal  places)  between 
the  mantissa  taken  and  the  next  larger. 


LOGARITHMS  303 

EXAMPLES    CXXI 

Find  from  the  table  the  logarithms  of  the  following  numbers: 
1.   798.                             2.   8.62.  3.   7784. 

4.   82.36.  5.   .8127.  6.  8516. 

7    78754.  8.   84.936.  9.   .077284. 

10.   8682.56.  11.   78.5716.  12.   .00859465. 

498.    To  find  from  the  table  a  number  whose  logarithm  is  given. 

We  find  in  the  table  the  next  lower  mantissa.  The  first  three 
figures  of  the  number  will  be  in  the  same  horizontal  line  in  the 
column  headed  N,  and  the  fourth  figure  at  the  top  of  the  column 
in  which  the  last  four  figures  of  the  next  lower  mantissa  are 
found.  The  remaining  figures  are  found  by  one  of  the  auxiliary- 
tables.  Thus,  to  find  x  when  log  x  =  2.904567,  we  proceed  as 
follows : 

The  next  lower  mantissa  is  .904553,  which  corresponds  to  the 
number  8027.  The  difference  between  this  mantissa  and  the  next 
higher  is  54  (in  the  last  two  decimal  places),  and  this  shows 
which  of  the  auxiliary  tables  to  use.  The  difference  between 
log  x  and  log  8027  is  14.  This  exact  difference  is  not  found  in  the 
auxiliary  table  headed  54,  but  the  next  lower  difference,  11,  gives 
2  for  the  5th  figure  of  the  number.  This  leaves  a  difference  of 
14  —  11  =  3  still  to  be  provided  for.  As  the  differences  for  the 
6th  figure  are  only  one  tenth  of  the  differences  in  the  auxiliary 
table,  we  multiply  our  difference  by  10  (this  is  the  same  as  divid- 
ing the  differences  in  the  auxiliary  table  by  10),  giving  30.  The 
next  smaller  difference  in  the  table  is  27,  which  gives  5  for  the 
6th  figure  of  the  number.  The  characteristic  2  shows  that 
the  number  has  three  figures  to  the  left  of  the  decimal  point. 
Hence  a  =  802.725. 

EXAMPLES  CXXH 

Find  from  the  table  the  numbers  of  which  the  following  are 
the  logarithms : 

1.  3.919758.  2.   2.899137. 

3.   0.912451.  4.   4.938343. 


304  HIGHER  ALGEBRA 

5.  1.902140.  6.  1.922995. 

7.   2.888596.,  8.  3.930272. 

9.   5.934953.  10.  5.913814. 

11.  3.903090.  12.  5.913578. 

Solve  the  following  by  logarithms : 

13.  7746  :  8334  ::  8027  :  x. 

14.  77.53  :  821.6  :  :  8.097  :  x. 

80.75  x  82.79  x  791.7 
8.692  x  770.4 

16.   1-v/791673000.  17.   8*  =  79846. 

18.   7.94s6  =  86978.  19.  8.6*  =  7896.24. 

Solution  of  17th.     Passing  to  logarithms,  we  have 
x  log  8  =  log  79846, 

.  log  79840     4.902253      r  ,oe„ 

whence  x  =  — p = =  5.4283« 

log  8  .90309 


CHAPTER   XXII 
INDETERMINATE   EQUATIONS  OF  THE   FIRST  DEGREE 

499.  An  Indeterminate  Equation  is  an  equation  having  two  or 
more  unknown  quantities,  this  equation  expressing  the  only  con- 
dition imposed  upon  the  unknown  quantities.     Hence, 

A  Set  of  Equations  is  Indeterminate  when  it  contains  more  un- 
known quantities  than  equations ;  for  by  elimination  the  set  may 
be  reduced  to  a  single  equation  containing  two  or  more  unknown 
quantities. 

As  shown  in  Art.  380,  the  unknown  quantities  of  such  an  equa- 
tion are  variables  and  admit  of  an  infinite  number  of  values. 

Thus,  in  the  equation  bx  —  3  y  =  11, 

we  may  give  to  one  of  the  variables  any  value  we  please  and  find  for  the  other 
such  a  value  as  will  satisfy  the  equation. 
Again,  from  the  equations 

3x-y  +  4z  =  22, 

4  x  +  3  y  -  2  z  =  19, 

we  have  by  eliminating  z,  11  x  +  by  =  60, 

in  which  x  and  y  admit  of  an  infinite  number  of  values. 

500.  A  single  equation  having  more  than  one  unknown  quantity 
is  indeterminate  in  a  less  general  sense,  if  an  additional  condition 
not  capable  of  being  expressed  in  an  equation  is  imposed. 

Thus,  let  it  be  required  to  find  the  positive  integral  values  of  x  and  y  in 

7x-f  6y=118. 

The  introduction  of  the  condition  that  the  values  are  to  be  positive  integers 
greatly  restricts  the  number.     In  this  equation 

when  x  =  4,    y  =  18, 

when  x  =  9,    y  =  1 1 , 

when  x  =  14,  y  =  4, 
downey's  alg.  — 20  305 


306  HIGHER  ALGEBRA 

and  it  may  be  shown  that  these  are  the  only  positive  integers  that  will  satisfy 
the  equation. 

The  discussion  here  will  be  confined  to  positive  integral  values  of  the 
unknown  quantities. 

501.  Any  equation  of  the  first  degree  containing  two  unknown 
quantities  can  be  reduced  to  one  of  the  forms  ax  ±  by  =  ±  c,  in 
which  a,  b,  c  are  positive  integers.  The  form  ax  +  by  =  —  c  is 
not  satisfied  for  any  positive  integral  values  of  x  and  y,  and  the 
form  ax  —  by  =  —  c  is  equivalent  to  by  —  ax  =  c ;  hence  we  need 
to  consider  only  the  forms  ax  ±  by  =  c. 

502.  Theorem.  The  forms  ax  ±by  —  c  have  no  positive  integral 
values  of  x  and  y,  if  a  and  b  have  a  common  factor  not  contained 
in  c. 

Dem.  Dividing  both  members  by  this  common  factor,  the 
second  member  is  fractional  for  all  values  of  x  and  y,  while  the 
first  member  is  integral  for  all  positive  integral  values  of  x  and  y ; 
hence  the  equations  are  not  satisfied  for  such  values. 

503.  Theorem.  The  number  of  positive  integral  values  of  x  and  y 
in  the  form  ax—by  =  c(a  and  b  being  prime  to  each  other)  is  limited. 

Dem.     Solving  for  x,  we  have 

c—  by 

x  — £• 

a 

Now  we  can  use  only  those  positive  integral  values  of  y  that 
will  render  c  —  by  positive  (i.e.,  by  >  c)  and  divisible  by  a  ;  hence 
the  number  of  positive  integral  values  of  both  x  and  y  is  limited. 

504.  Theorem.  The  number  of  positive  integral  values  of  x  and 
y  in  the  form  ax  —  by  =  c  (a  and  b  being  prime  to  each  other)  is 
infinite. 

Dem.     Solving  for  x,  we  have 

c  +  by 

x  = — 

a 

No  positive  integral  value  of  y  will  render  x  negative,  and  an 
infinite  number  of  such  values  will  render  c  +  by  divisible  by  a ; 
hence  the  number  of  positive  integral  values  of  both  x  and  y  is 
infinite. 


INDETERMINATE  EQUATIONS  307 


EXAMPLES  CXXIH 

Find  the  positive  integral  values  of  x  and  y  in  each  of  the 
following : 

1.  5»  +  9y  =  37. 

Solution.     Solving  for  x  in  terms  of  y,  we  have 

•     Aii'fi^Jt  (!) 

5 
Since  x  must  be  positive, 

9*/>37, 

whence  2/  >  4$, 

and  as  ?/  must  be  integral,  it  cannot  be  greater  than  4. 
Now  (1)  may  be  written  in  the  form 

x  =  7-v  +  l^A£  =  7-,J  +  2±zl2.  (2) 

o  5 

To  make  x  integral  *•  must  be  integral.     The  only  value  of  y  not 

5 

i  o  ft 

greater  than  4  that  will  make ^  integral  is  3.    This  substituted  in  either 

5 
(1)  or  (2)  gives  x  —  2. 

2.  8a; +  13  7/ =138. 

Solution.     Solving  for  x  in  terms  of  y,  we  have 

x  =  ™-1*'.  (1) 

8  v  J 

Since  x  must  be  positive, 

13  y  >  138, 

whence  y  >>  10. 

Now  (1)  may  be  written  in  the  form 

x  =  n-y+'l^JL.  (2) 

When  it  is  not  easy  to  see  by  inspection  all  the  values  of  y 
that  will  render  the  fractional  part  integral,  we  may  proceed  as 

follows : 

2 5  « 

Since    — ^   must  be  an  integer,  any  integral  number  of  times  this 

8 
quantity  will  be  an  integer.     Let  us  multiply  by  the  smallest  number  that 
will  give  a  remainder  1   (or   -  1)  on  dividing  the  coefficient  of  y  by  the 
denominator.    This  multiplier  in  this  case  is  5,  and  we  have 


308  HIGHER  ALGEBRA 

It  is  now  easy  to  see  that  2  and  10  are  the  only  two  numbers  not  greater 

an  10  tl 
14  and  1. 
Or 

givini 


than  10  that  will  make  -  integral,  and  the  corresponding  values  of  x  are 

8 

Or,  if  we  choose,  we  may  represent     ~  y  by  the  undetermined  integer  ra, 


=  m, 


8 
whence  y  =  2  -  8  m,     •  (4) 

and  from  (1)  x  =  14  +  13  m.  (5) 

It  is  seen  that  x  and  y  are  integral  for  all  integral  values  of  m,  and  (4)  and 
(5)  constitute  what  is  called  the  general  solution  in  integers  of  the  given 
equation  ;  but  (4)  shows  that  only  0  and  negative  values  of  m  will  give  posi- 
tive values  for  y,  and  (5)  shows  that  m  cannot  be  smaller  than  —  1.  Hence 
0  and  —  1  are  the  only  admissible  values  of  m. 

Now  when  m  =  0,       x  =  14,  and  y  =  2  ; 

and  when  m  =  —  1,    x  =  1,    and  y  =  10. 

3.  13a -30?/  =  61. 

Solution.  From  Art.  504  we  know  that  x  and  y  have  an  infinite  number 
of  values.     Solving  for  x  in  terms  of  y,  we  have 

g  =  6i+aoy  =  4  +  2    +o±±g. 

13  *  13 

Multiplying  the  numerator  of  the  last  fraction  by  10  and  reducing  to  a 
mixed  quantity,  we  have 

<*>  +  40y  =  64.3y  +  i2+x 

13  J  13 

The  values  of  y  that  make  12  +  y  integral  are  I,  14,  25,  etc.,  and  the  cor^ 
13 

responding  values  of  x  are  7,  37,  67,  etc. 

4.  14#-21?y  =  32. 

As  proved  in  Art.  502,  the  solution  is  impossible. 

5.  3z  +  7#  =  10.  6.  3x  +  5y  =  26. 
7.  5x-Uy=ll.  8.  2»-M#  =  10. 
9.   27 x  +  18?/  =  47.                         10.  4a;- 19?/ =  23. 

11.   3#  +  7?/  =  58.  12.   13a  +  2?/  =  119. 

13.   14#-5?/  =  7.  14.   2Sx-Soy  =  23. 

3?/-5z  =  -8,  f20a-21?y  =  38, 


15. 


|2a;  +  3?/-52  =  -8J  {20*-- 

\5a>-y  +  43  =  21.  j  3?/  +  4 


2=34. 


INDETERMINATE  EQUATIONS  309 

PROBLEMS   LEADING  TO   INDETERMINATE    EQUATIONS 
EXAMPLES  CXXIV 

1.  A  man  employed  two  squads  of  men,  paying  $3a  day  to 
each  of  the  first  squad  and  $4  a  day  to  each  of  the  second.  He 
paid  them  all  $  06  a  day.     How  many  belonged  to  each  squad  ? 

2.  A  dairyman  paid  $  752  for  cows,  $  37  each  for  Jerseys  and 
$  23  each  for  Durhams.     How  many  of  each  did  he  buy  ? 

3.  In  how  many  ways  can  a  debt  of  $  43  be  paid  with  2-dollar 
and  5-dollar  bills  ? 

4.  A  man  invested  $  10,000  in  town  lots,  paying  $  190  each 
for  those  in  one  locality,  and  $  130  each  for  those  in  another. 
Find  the  number  in  each  locality. 

5.  In  how  many  ways  can  a  debt  of  £  50  be  discharged  with 
guineas  and  3-shilling  pieces  ?     £  51  ? 

6.  What  fraction  becomes  J  when  its  numerator  is  doubled 
and  its  denominator  is  increased  by  7  ? 

7.  How  many  4-pound  and  6-pound  weights  must  be  used  to 
weigh  45  pounds? 

8.  Divide  136  into  two  parts,  one  of  which  when  divided  by 
5  leaves  a  remainder  2,  and  the  other  divided  by  8  leaves  a 
remainder  3. 

9.  How  many  times  each  must  a  7-inch  stick  and  a  13-inch 
stick  be  applied  to  measure  4  feet  ? 

10.  A  owes  B  a  shilling.  A  has  only  sovereigns,  and  B  has  only 
dollars,  worth  4s.  3d.  each.     How  can  A  most  easily  pay  B  ? 

11.  A  wholesale  grocer  received  $  180  for  15  barrels  of  sirup, 
the  prices  being  $  10,  $  11,  and  $  13  per  barrel.  How  many  of 
each  kind  did  he  sell  ? 

12.  A  farmer  paid  $  160  for  pigs,  sheep,  and  calves.  The  pigs 
cost  |  3  each,  the  sheep  $  4  each,  and  the  calves  $  7  each  ;  and  the 
number  of  calves  was  equal  to  the  number  of  pigs  and  sheep 
together.     How  many  of  each  did  he  buy? 


CHAPTER   XXIII 

THEORY    OF    EQUATIONS    AND    SOLUTION    OF    NUMERICAL 
HIGHER  EQUATIONS 

505.    When  a  quadratic  equation  has  the  general  form 
x2+px  =  q, 
we  have  seen  (Art.  344)  that 


-f±Vf+* 


in  which  the  unknown  quantity  is  expressed  in  terms  of  the 
general  coefficients,  giving  a  formula  by  which  it  may  be  found 
in  all  special  cases.  Solutions  of  general  (i.e.,  literal)  equations 
of  the  third  and  fourth  degree  are  known  (see  the  methods  of 
Cardan  and  Descartes,  Arts.  573  and  576) ;  but  these  solutions 
have  in  many  cases  difficulties  which  render  them  of  little  practi- 
cal value,  and  they  are  seldom  employed.  For  general  equations 
above  the  fourth  degree  no  solution  is  known,  i.e.,  no  method  of 
expressing  in  terms  of  the  literal  coefficients,  the  values  of  the 
unknown  quantity,  —  indeed,  it  has  been  shown  by  Abel,  a  Nor- 
wegian mathematician,  that  it  is  impossible  thus  to  express, 
algebraically,  the  values  of  the  unknown  quantity.  So  soon, 
however,  as  numerical  values  are  assigned  to  the  coefficients,  the 
real  roots  may  in  all  cases  readily  be  found,  either  exactly  or  to 
any  required  degree  of  approximation. 

506.  A  Root  of  an  Equation  is  a  quantity  which,  substituted  for 
the  unknown  quantity,  satisfies  the  equation.  It  may  be  com- 
mensurable, incommensurable,  or  imaginary. 

507.  Commensurable  Numbers  are  such  as  can  be  exactly  ex- 
pressed in  the  decimal  notation.      They  are  integers,  common 

310 


-REDUCTION   TO   THE  NORMAL  FORM  311 

fractions,  terminating  decimals,  and  repeating  decimals  (the 
latter,  as  shown  in  ex.  25,  page  205,  being  reducible  to  equivalent 
common  fractions). 

508.  Incommensurable  Numbers  are  such  as  cannot  be  exactly- 
expressed  in  the  decimal  notation. 

Thus,  V2,  2  ±  V3,   y/E  are  incommensurable  numbers. 

509.  A  Multiple  Root  is  a  root  that  occurs  more  than  once  in  an 
equation.  If  it  occurs  twice,  it  is  called  a  double  root;  if  three 
times,  a  triple  root;  if  four  times,  a  quadruple  root,  etc. 

Thus,  from  x5  -  4  x4  +  x3  -f  10  x2  -  4  x  -  8  =  0,  we  have  (Art.  101) 
(X  _2)(x  -  2)(x  -  2)(x  +  l)(x  +  1)=  0, 
giving  (Art.  348)  2  as  a  triple  root  and  —  1  as  a  double  root. 

We  may  say  that  an  equation  has  the  same  root  repeated  several 
times  and  call  it  a  Multiple  Root,  or  that  it  has  several  roots  of 
the  same  value  and  call  them  Equal  Roots. 

510.  An  equation  is  said  to  be  in  the  Normal  or  Typical  Form 
when  the  exponents  are  all  positive  integers,  the  coefficient  of  the 
first  term  (highest  power)  is  1,  and  the  other  coefficients  are  all 
integers. 

REDUCTION  TO  THE  NORMAL  FORM 

511.  Theorem.  Every  algebraical  equation  having  rational  co- 
efficients can  be  transformed  into  an  equation  of  the  normal  form  in 
terms  of  a  new  unknown  quantity  ivhich  is  a  known  function  of  the 
original  unknown  quantity. 

Dem.  1st.  To  make  the  exponents  positive,  the  equation  is  mul~ 
tiplied  by  the  unknown  quantity  ivith  a  positive  exponent  equal 
numerically  to  the  largest  negative  exponent.  This  neither  destroys 
the  equality  of  the  members  nor  changes  the  value  of  the  un- 
known quantity. 

2d.  To  make  the  exjjonents  integral,  each  exponent  is  multiplied 
by  the  1.  c.  m.  of  the  denominators  of  the  fractional  exponents,  and 
the  unknown  quantify  is  rrj>hici>d  by  another.  If  the  equation  is  in 
x,  and  if  m  is  the  1.  c.  m.  of  the  denominators  of  the  fractional 


312  HIGHER  ALGEBRA 

exponents,  the  transformation  consists  simply  in  substituting  ym 
for  x.  While  this  changes  the  roots,  the  roots  of  the  original 
equation  are  known  functions  of  the  roots  of  the  transformed 
equation,  viz.,  the  mth  power  of  them. 

3d.  To  make  the  coefficients  integral,  and  the  first  one,  that  of  yn, 
unity,  we  divide  the  equation  by  the  coefficient  of  yn,  multiply  the 
coefficient  of  yn~l  by  k,  that  of  yn~2  by  k2,  that  of  yn~3  by  fc8,  and  so 
on,  replacing  y  by  z,  and  then  take  k  of  such  value  as  will  make  each 
coefficient  integral.  After  the  operations  of  1st  and  2d,  and  divid- 
ing by  the  coefficient  of  the  highest  power,  the  equation,  if 
arranged,  has  the  form 


r+^2/n-1  +  ^r-2  +  ^2/n-3  +  " 

..)-* 

Substituting  y  =  -,  this  becomes 
k 

zn      a  zn  l      b  zn~2      c  2n-3 
kn     a'  kn~l     b'  kn~2     c' kn3 

4-» 

Multiplying  by  kn, 

(1) 


„  ,  ak    ■   ,   ,  bk2  n  9  ,  cZc3   „  ,  ,        lkn      _  /ox 

Z  +'a1  Z      +  "F Z      +7r  Z      +  '"T  =  °-  (2) 

Such  value  of  k  may  now  be  taken  (it  should  be  the  least 
possible)  as  will  remove  the  denominators. 

z 


We  now  have  x  =  ym  =  f  -  j  , 

and  the  roots  of  the  original  equation  become  known  when  the 
roots  of  the  equation  in  z  become  known. 

512.  Cor.  From  (1)  and  (2)  it  is  seen  that  to  obtain  an  equation 
whose  roots  are  k  times  those  of  a  given  equation  we  have  but  to 
multiply  the  coefficient  of  the  2d  term  by  k,  that  of  the  3d  by  k2, 
that  of  the  kth  by  Zc3,  and  so  on  to  the  absolute  term,  which  is 
multiplied  by  kn. 

EXAMPLES    CXXV 

Transform  the  following  into  equations  of  the  normal  form  and 
express  x  in  terms  of  the  new  unknown  quantity : 


TEST  FOR   ROOTS  313 

1.   12  a*  -  3  af *  -  8  a*  +  10  a;*  -  3  af*  -  10  =  0. 
Solution.     Multiplying  by  x7, 

12  x^  -  3  x$  -  8  x*  +  10  x*  -  3  -  10  x$  =  0. 
Multiplying  the  exponents  by  6  and  replacing  x  by  y, 

12  y*  -  3  y  -  8  y5  +  10  y3  -  3  -  10  y2  =  0. 
Arranging  and  dividing  by  —  8, 

y5  -  |  y*  -  |  y3  +  |  y2  +  f  */  +  f  =  o. 

Introducing  A:  as  indicated  in  the  3d  part  of  the  demonstration  and  replac- 


ing y  by  z, 

Making  k  =  2 


^_3^^_^  3     5F  2     3^        3^_ 


^5  _  3  Zi  _  5  ^  +  10  z*  +  6  0  +  12  =  0, 


in  which 


*=H!)6 


By  Art.  363,  it  is  found  that  —  2  is  a  root  of  this  equation  ;  hence 

y  =  =£■  =  - \,  and  x=(-  1)«  =  1. 
Therefore  1  is  the  corresponding  root  of  the  original  equation. 

4.  }~*t=***  +  *-  5.  2Vr^  =  4-3af* 

1  +  #  _1       jc  +  2 

6.   3ar5-2x4-?ar3  +  5^--^  +  1  =  0. 

5  ar     o  a;     o         10 

TEST  FOR  ROOTS 

513.  Theorem.  If  f(x)  be  divided  by  x  —  a,  the  last  remainder 
irill  be  f(a);  i.e.,  will  be  what  f(x)  becomes  when  a  is  substituted 
for  x. 

Dem.  Let  <f>(x)  be  the  quotient  and  r  the  last  remainder. 
Then,  by  the  laws  of  division, 

/(*)«(«-a)f(*)  +  r. 


314  HIGHER  ALGEBRA 

As  this  equation  is  true  for  all  values  of  x,  it  is  true  when 
x  =  a.     Substituting  a  for  x,  we  have 

/(a)=r. 

514.  Sch.  It  follows  from  the  last  theorem  that  the  short 
method  of  dividing  by  x  —  a,  given  in  Art.  79,  becomes  a  short 
method  of  substituting  any  number  for  x. 

The  operation  of  substituting  a  value  for  x  is  often  called 
evaluating. 

EXAMPLES   CXXVI 

Find  what  the  following  become  when  the  indicated  values  of  x 
are  substituted : 

1-  x5-2x4-4:Xi  +  3x2-5x  +  6  for  x  =  3. 
Solution.     By  the  method  of  division  given  in  Art.  79,  we  have 
x5-2x4-4x3  +  3x2-5x  +  6  [_3 
1      _ i  o      -5-9 

Therefore,  by  Art.  513,  this  polynomial  becomes  —  9  when  3  is  substi- 
tuted for  x. 

2.  x*  +  a?  ~  17  x2  -  10  x  +  11  for  x  =  4. 

3.  ^-8^  +  18^-12^-14^-4  for  x  =  5. 

4.  3a6-4af-6ar5-4arJ-10;e  +  23  for  x  =  2. 

Operation 
3  x6  -  4  x5  +  0  x4  -  6  x3  -  4  x2  -  10  x  +  23  |_2 
2  4  2  0-10  3 

Therefore  this  polynomial  becomes  3  when  2  is  substituted  for  x.  The 
same  operation  shows  (Art.  79)  that  when  the  polynomial  is  divided  by 
x  —  2,  the  quotient  is  3  x5  +  2  x4  +  4  x3  +  2  x2  —  10,  with  a  remainder  of  3. 

5.  5  x5  +  12  x4  +  6  x3  -  3  a;2  -  13  x  +  9  for  a;  =  -  2. 

6.  4  a5  -  17  x4  +  18  x2  -  20  for  z  =  4. 

7.  x6  -  8  x*  +  10  z4  +  40  x3  -  71  x2  -  32  a  +  60  for  x  =  5. 

515.  Theorem.     7/"  /(.r)   &  divisible   by  x  —  a,  a  is  a  rootf   o/ 
.  f(x)  =  0 ;    and,  conversely,   if  a  is  a  root  of  f(x)  =  0,  f(x)  is 

divisible  by  x  —  a. 


NUMBER  AND   CHARACTER   OF  ROOTS  315 

Dem.  By  Art.  513  if  f(x)  be  divided  by  x  —  a,  the  remainder 
is  /(a).  Now,  if  the  remainder  is  0,  /(a)  =  0,  and  the  equation 
is  satisfied. 

Again,  using  the  notation  of  Art.  513, 

'(x  —  a)+(x)  4-  r  =/(*). 

Now  if  a  is  a  root,  f(x)  is  0  and  x  —  a  is  0,  and  the  equation 
becomes  0  +  r  =  0. 

Hence,  the  remainder  being  0,  f(x)  is  divisible  by  x  —  a. 

516.  Cor.  If  a  is  a  root  of  f(x)  =  0,  a  is  a  factor  of  the  abso- 
lute term. 

If  a  is  a  root,  there  is  no  remainder  when  f(x)  is  divided  by 
x  —  a ;  therefore,  as  readily  appears  from  the  process  of  division, 
a  is  a  factor  of  the  absolute  term. 

NUMBER  AND  CHARACTER  OF  THE  ROOTS 

517.  Theorem.  An  equation  in  the  normal  form  cannot  have  a 
fractional  root,  and  hence  the  real  roots  of  such  an  equation  are 
either  integral  or  incommensurable. 

Dem.  Suppose  -,  a  simple  fraction  in  its  lowest  terms,  to  be  a 
root  of  the  normal  equation 

xn  +  Axn~l  +  Bxn  2  4-  Cxn-S  +  •••  L  =  0. 
Substituting  this  value  of  x,  we  have 


+  -  +  £  =  0. 


In  this  equation  all  the  terms  except  the  first  are  integral,  and 
the  first  is  a  simple  fraction  in  its  lowest  terms,  s  and  t  being 
prime  to  each  other.  But  the  sum  of  a  simple  fraction  in  its 
lowest  terms  and  a  series  of   integers  cannot   be  0.     Hence  x 

cannot  equal  -,  a  fraction. 

518.  Sch.  It  is  readily  seen  that  the  above  reasoning  does  not 
apply  if  the  coeificient  of  the  first  term  is  other  than  1 ;  for  in 


316  HIGHER  ALGEBRA 

that  case  tlie  coefficient  might  contain  t,  making  the  first  term 
integral  instead  of  fractional  when  -  is  substituted  for  x. 

519.  Theorem.  An  equation  of  the  nth  degree  has  n  roots  and  no 
more* 

Dem.  Let  a  be  a  root  of  /(a?)  =  0.  Then  f(x)  is  divisible  by 
x  —  a  (Art.  515).     Representing  the  quotient  by  <f>(x),  we  have 

O-a)<£(»=0.  (1) 

This  is  satisfied  not  only  for  x  —  a  =  0,  but  also  for  <f>(x)  =  0 
(Art.  348),  and  the  degree  of  <f>(x)  is  one  lower  than  that  of  fix). 
If  b  is  a  root  of  <f>(x)  =  0,  it  is  also  a  root  of  f(x)  =  0,  as  seen 
from  (1).  Dividing  <f>(x)  by  x  —  a,  the  degree  will  again  be 
diminished  by  1.  This  process  may  be  continued  for  n  divisions 
and  no  more.     Hence  f(x)  =  0  has  n  roots  and  no  more. 

520.  Cor.  i.  To  form  an  equation  ivhose  roots  shall  be  a,  b,  c, 
•  ••  I,  we  have 

f(x)  =  (x  —  a)  (x  —  b)  (x  —  c)  •  •  •  (x  —  l)  =  0 ; 

i.e.,  to  form  an  equation  having  any  roots  whatever,  ive  have  but  to 
subtract  these  roots  from  x  and  place  the  product  of  the  remainders 
equal  to  0. 

521.  Cor.  2.  The  equation  f(x)  =  0  may  have  2,  3,  or  even  n 
equal  roots,  as  in  the  above  demonstration  we  may  have 

a  =  b,  a  =  b  =  c,  or  a  —  b  =  c  =  •  •  •  I. 

522.  Theorem.  If  an  equation  having  only  rational  coefficients 
has  imaginary  or  quadratic  surd  roots,  these  roots  enter  in  conjugate 
pairs  (Art.  221). 

Dem.  If  /3 V  —  1  is  a  root  of  f(x)  =  0,  x  —  /?V—  1  is  a  factor 
of  f(x)  (Art.  515).  Now  the  only  factor  by  which  x  —  /?V—  1 
can  be  multiplied  to  produce  a  real  and  rational  quantity  is 
x  +  fi-y/—  1,  giving  x2  +  /32  for  the  product.  Hence  (Art.  515) 
—  /?V—  1  is  also  a  root. 

*  It  is  here  assumed  that  every  equation  has  at  least  one  root.  This  has 
been  proved  by  Argand,  Gauss,  Cauchy,  Clifford,  and  others. 


NUMBER  AND  CHARACTER   OF  ROOTS  317 

Again,  if  a  -+-  /?V—  1  is  a  root  of  f(x)  =  0,  x  —  (a-{-  /?V—  1) 
is  a  factor  of  f(x) ;  hence  to  give  a  real  and  rational  product, 
x  —  (a  —  /JV^T)  must  also  be  a  factor,  giving  «  —  /?V—  1  as  a 
root,  and  (#  —  «)2  +  /32  as  the  product  of  the  two  factors. 

It  may  be  shown  in  the  same  way  that  if  /?Vy  is  a  root  of 
f(x)  =  0,  —  ftVy  must  also  be  a  root;  and  that  if  a  +  /?Vy  is  a 
root  of  /(#)  =  0,  a  —  /?  Vy  must  also  be  a  root. 

523.  Cor.  i.  An  equation  of  odd  degree  has  at  least  one  real 
root. 

524.  Cor.  2.  All  the  roots  of  an  equation  of  even  degree  can  be 
imaginary  only  when  the  absolute  term  is  positive. 

For,  as  seen  above,  the  product  of  the  factors  obtained  by  sub- 
tracting conjugate  imaginaries  from  x  always  gives  a  positive 
absolute  term. 

525.  Cor.  3.  An  equation  of  even  degree  whose  absolute  term  is 
negative  has  at  least  two  real  roots,  and  these  have  opposite  signs. 

526.  Theorem.  Changing  the  signs  of  the  terms  containing  the 
odd  powers,  or  the  even  powers,  or,  if  the  equation  is  complete,  the 
alternate  terms,  changes  the  signs  of  the  roots  of  an  equation;  i.e., 
the  positive  roots  of  f{—  x)  =  0  are  the  negative  roots  of  f(x)  =  0. 

Dem.  Let  a,  b,  c,  •••/  be  the  roots  of  an  equation  f(x)  =  0; 
then  (Art.  520) 

f(x)  =  (x-  a)(x  -b)(x-  c)  ...  (x  -  I)  =  0. 

Replacing  x  by  —  x,  we  have 

/(-  x)  =  (-  x  -  a)  (-  x  -  b)  (-  x  -  c)  ...  (-  x  -  I)  =  0, 

or,  removing  —  1  from  each  factor, 

/(-  x)  =  (-  l)n(x  +  a)  (x  +  b) (p  +  e)  -  (a?  + 1)?* 0. 

By  equating  each  of  these  binomial  factors  with  0  (Art.  348), 
the  roots  are  seen  to  be  —  a,  —  b,  —  c,  •••  —  I;  i.e.,  changing  the 
sign  of  x  changes  the  signs  of  the  roots.  But  changing  the  sign 
of  x  changes  the  signs  of  the  terms  containing  the  odd  powers, 
and  only  these. 


318  HIGHER   ALGEBRA 

Changing  the  signs  of  the  terms  containing  the  even  powers 
also  changes  the  signs  of  the  roots,  since  this  is  the  same  as 
multiplying  or  dividing  both  members  by  —  1  after  changing  the 
signs  of  the  terms  containing  the  odd  powers. 

527.  Sch.  The  absolute  term  may  be  regarded  as  the  coeffi- 
cient of  x°,  and  it  must  be  noted  that  0  is  an  even  number. 

528.  A  Permanence  is  a  succession  of  two  like  signs  in  the  con- 
secutive terms  of  a  polynomial ;  a  Variation,  of  two  unlike  signs. 

Thus,  in  x6  —  2  xs  —  4  x*  +  5  x3  +  3  x2  +  x  —  4,    the   signs  being    -j 

+  +  H — ,  there  are  three  permanences  and  three  variations. 

529.  Descartes'  Rule  of  Signs.  An  equation  in  the  form  f(x)  =  0, 
ivJiether  complete  or  incomplete,  cannot  have  more  positive  roots  than 
the  number  of  variations  in  f(x),  nor  more  negative  roots  than  the 
number  of  variations  in  f(—x). 

Verification.     Let  the  signs  and  missing  terms  of  f(x)  be 

-f +0-+00 +, 

giving  4  variations. 

Let  us  now  introduce  a  positive  root  by  multiplying  by  x  minus 

this  root  (Art.  515),  writing  only  the  signs  in  the  operation,  as 

follows : 

+  +0-+00 + 


+  +0-+00 + 

0+-00+  +  + 

+  ± +-0-TT   + 


The  ambiguous  signs  ±  and  q=  are  placed  where  the  sign  to  be 
used  must  be  determined  by  the  relative  numerical  values  of  the 
coefficients. 

Counting,  now,  the  fewest  number  of  variations  that  can  occur, 
by  taking  that  one  of  the  ambiguous  signs  that  will  give,  with  the 
preceding  sign,  a  permanence,  we  have  5  variations,  a  variation 
being  necessarily  introduced  at  the  end.  Thus  each  time  a  pos- 
itive root  is  introduced  by  multiplying  by  x  minus  that  root,  at 
least  one  variation  is  added.  Hence  an  equation  cannot  have 
more  positive  roots  than  the  number  of  variations. 


NUMBER  AND   CHARACTER   OF  ROOTS  319 

Again  the  positive  roots  of  /(—  x)=  0  are  the  negative  roots  of 
f(x)=Q  (Art.  526).  As  just  shown, /(—  x)  =  0  cannot  have  more 
positive  roots  than  the  number  of  variations  in  /(—  x).  Hence 
f(x)  sa  0  cannot  have  more  negative  roots  than  the  number  of 
variations  in  /(—#). 

530.  Cor.  i.  An  equation  whose  terms  are  all  positive  has  no 
positive  roots. 

531.  Cor.  2.  An  equation  whose  terms  of  ev&n  power  are  all  of 
one  sign  and  ichose  terms  of  odd  power  are  all  of  the  contrary  sign 
has  no  negative  roots.      Why? 

Hence  a  complete  equation  whose  terms  are  alternately  +  and  — 
has  no  negative  roots. 

532.  Cor.  3.  An  equation  having  only  even  powers,  and  these  all  of 
the  same  sign,  has  no  real  roots.  An  equation  having  only  odd 
powers,  and  these  all  of  the  same  sign,  has  no  real  root  except  0. 

533.  Note.  The  truth  of  the  above  corollaries  is  readily  seen 
independently  of  Descartes'  Rule  of  Signs. 

534.  Cor.  4.  An  incomplete  equation  has  at  least  as  many  imagi- 
nary roots  as  the  difference  between  the  degree  of  the  equation  and  the 
total  number  of  variations  in  f(x)  and  f(—x). 

For  the  total  number  of  roots  is  the  same  as  the  degree  of  the 
equation,  and  there  cannot  be  more  real  roots  than  the  total  num- 
ber of  variations  in  f(x)  and  /(—  x). 

535.  Cor.  5.  A  complete  equation  cannot  have  more  negative  roots 
than  its  number  of  permanences. 

The  total  number  of  variations  and  permanences  is  one  less  than 
the  number  of  terms,  i.e.,  the  same  as  the  degree  of  the  equation 
or  the  total  number  of  roots.  Now  there  cannot  be  more  positive 
roots  than  the  number  of  variations,  and  hence  not  more  negative 
roots  than  the  number  of  permanences. 

536.  Note.  Descartes'  Rule  and  the  corollaries  deduced  from 
it  are  useful  only  in  preventing,  in  some  cases,  a  fruitless  search 
for  roots  of  a  particular  sign  after  all  of  that  sign  have  been 
found,  or  for  the  situation  of  roots  after  the  situation  of  all  the 
real  ones  has  been  found. 


320  HIGHER  ALGEBRA 

EXAMPLES   CXXVII 

Determine  the  greatest  number  of  positive  and  negative,  and 
the  least  number  of  imaginary,  roots  in  the  following : 

1.  »6  +  8ar3  +  3oj-6  =  0. 

Solution.  By  Art.  525  this  equation  has  2  real  roots,  1  positive  and  1 
negative.  Since  there  is  but  1  variation,  the  equation  cannot  have  more  than 
1  positive  root.  Changing  the  signs  of  the  terms  containing  the  odd  powers 
of  «,  we  have 

xe  _  8  x3  -  3  x  -  6  =  0. 

Since  this  has  but  1  variation,  the  original  equation  cannot  have  more  than 
1  negative  root.  But  the  equation,  being  of  the  6th  degree,  has  6  roots  ; 
hence  it  has  4  imaginary  roots. 

2.  a;4  +  12a2  +  5a -10  =  0. 

3.  3^  +  4^  +  5=:  0. 

4.  ar5  -  7  x2  -  G  =  0. 

5.  x6-  7^-^  +  3  =  0. 

6.  2a6-435-5<B8  +  3a2-5  =  0. 

7.  x«  +  7x5-4:X4-5xi-3x-6  =  0. 
a  xc,-2x*  +  x4-x2-l  =  0. 

9.   xn  —  1,  when  n  is  odd.  10.   xn  —  1,  when  n  is  even. 

11.   xn  +  1,  when  n  is  odd.  12.   xn  +  1,  when  n  is  even. 

SOLUTION  FOR   COMMENSURABLE   ROOTS 

537.  The  process  of  finding  the  roots  of  an  equation  when  all 
the  roots,  or  all  but  two  of  them,  are  integral,  has  been  fully 
explained  in  Chapter  XVI.  We  now  see  from  Art.  513  that  the 
short  process  of  factoring  there  used  is  also  a  short  process  of 
evaluating  for,  or  substituting,  any  particular  value  of  the  unknown 
quantity.  A  few  additional  examples  are  here  given  for  further 
practice  in  that  important  process.  In  applying  it  the  student 
should  keep  in  mind  the  following  : 

1.  When  the  equation  is  in  the  normal  form,  only  factors  of 
the  absolute  term  need  be  tried  (Art.  516). 


SOLUTION  FOR   COMMENSURABLE  HOOTS  321 

2.  It  will  be  found  expedient  to  try  the  small- positive  factors 
first.  Before  writing  down  the  various  sums,  it  is  best,  if  the 
coefficients  are  not  too  large,  to  run  through  mentally  with  a 
factor  of  the  last  term  and  see  whether  the  last  addition  gives  0. 

3.  If  all  the  terms  are  positive,  no  positive  numbers  need  be 
tried  (Art.  102)  ;  and  if  at  any  stage  in  the  operation  of  evaluat- 
ing, all  the  sums  become  positive,  no  more  positive  numbers  need 
be  tried. 

4.  When  all  but  two  of  the  roots  have  been  found,  the  remain- 
ing two,  whatever  their  values,  may  be  found  by  solving  the 
resulting  quadratic.  The  last  four  roots,  even  if  surd  or  imagi- 
nary, may  be  found  if  the  alternate  terms  of  the  depressed  equa- 
tion are  0,  since  this  depressed  equation  would  then  have  the 
quadratic  form  (Art.  359). 

EXAMPLES  CXXVm 

Find  all  the  roots  of  the  following  : 

1.   a°-4ari-6a;4  +  40ar?-31arJ-36a:  +  36  =  0. 

Operation 
s6  _  4  ^5  _    e  a*  +  40  a*  -  31  x*  -  36  x  +  30  |  1 


-3 

-    9 

31 

0 

-36 

0|_2 

-1 

-11 

9 

18 

0 

[1 

1 

-    9 

-    9 

0 

Li 

4 

3 

0 

l-i 

3 

0 

|-3 

0 
Or  the  last  two  roots  may  be  found  by  solving  the  quadratic 
x2  +  4  x  +  3  =  0,  giving  x  =  -2±l=-l  and  -  3. 

2.   ^-3^-14^  +  24  =  0.  3.   x5-7x  +  6  =  0. 

4.  ar3-15ar>  +  74a;-120  =  0.     5.  ^-7^-52  +  35  =  0. 

6.  z4-10z3  +  24z2  +  10x-25  =  0. 

7.  a*4  -  10a?  +  20a?  +  10a; -21=0. 

a  x5  -  5x*  -  13ar>  +  Gox2  +  36x-  180  =  0. 
downey's  alg. — 21 


322  HIGHER   ALGEBRA 

9.  af  +  2x4-3a?-3x2  +  2x  +  l  =  0. 

10.  x5  -  6  x"  +  15  a?  -  26  x2  +  36  a?  -  24  =  0. 

11.  ar5  +  2  a;4  -  9  Xs  +  14  a;  -  8  =  0. 

12.  ar5  -  5  ar3  +  6  x2  -  14  a?  +  12  =  0. 

13.  x5  -  l&x4  +  85  a3  -  225  x2  +  274a>-  120  =  0. 

14.  af  -  2  x4  -  6  a;8  +  12  x2  +  9  a;  -  18  =  0. 

15.  ar5  —  x4  +  4  a?3  —  4  x2  +  4  a;  —  4  =  0. 

16.  4  ar5  -  4  x4  -  31  ar5  +  10  a;2  +  39  x  -  18  =  0. 

Suggestion.  When  the  coefficient  of  the  first  term  is  not  1,  first  find  all 
the  integral  roots,  and  then,  if  necessary,  transform  the  depressed  equation 
to  the  normal  form. 

17.  6  x5  -  19  x4  +  13  Xs  +  13  x2  -  19  x  +  6  =  0. 

18.  a;6  -  3  x>  -  3  x4  +  15  a?  -  6  a*2  -  12  x  +  8  =  0. 

19.  xc>  -  4  a^5  -  4  x4  +  20  Xs  -  x2  -  16  x  +  4  =  0. 

20.  a;7  -  2  a;6  -  10  a-5  +  28  x4  +  5  ar5  -  74  a;2  +  76  a?  -  24  =  0. 

21.  12  x6  -  20  x5  -  85  a4  +  92  x>  +  97  a;2  -  132  x  +  36  =  0. 

EQUATIONS   WITH  ONLY   EVEN  OR   ONLY   ODD  POWERS 

538.  When 'an  equation  contains  only  even  powers  of  x,  we  may- 
regard  x2  as  the  unknown  quantity  and,  without  supplying  the 
missing  odd  powers,  proceed  as  above  and  then  extract  the  square 
root  of  the  numbers  which  reduce  the  function  to  0.  This  is  of 
great  importance  in  finding  the  roots  of  such  an  equation  when 
they  are  monomial  surds  or  monomial  imaginaries. 

An  equation  containing  only  odd  powers  can,  after  dividing  by 
x  (for  in  that  case  there  could  be  no  absolute  term),  be  treated  in 
the  same  way. 

EXAMPLES  CXXIX 

Find  all  of  the  roots  of  the  following : 

1.   a6-14x4-f  49o2-36  =  0. 

Solution.    This  may  be  written 

(a2)8  -  14  (z2)2  +  49 (a2)  -  36  =  0. 


RELATION  OF  ROOTS   TO   COEFFICIENTS  323 

Regarding  x2  as  the  unknown  quantity,  there  are  no  missing  terms.     We 
may,  therefore,  proceed  as  follows  : 

(x2)3  -  14 (x2)2  +  49(x2) -  36  \_l 
-13  36  0  |_4 

-    9  0  |_9 

0 
Hence  the  factors  are  x2  — 1,  x2— 4,  x2— 9,  and  the  roots  are  ±  1,  ±2,  ±3. 

2.  x*-3x«-15xA  +  19^  +  30  =  0. 

Operation 

x*  -  3X6  -  15  x4  +  19x2  +  30  | 2 


1 

-17 

-15 

0  ( 5 

4 

3 

0 

L=i 

3 

|-3 

0 

Hence  the  factors  are  x2  —  2,  x2  —  5,  x2  +  1,  x2  +  3,  and  the  roots  are 
±V2,  ±V5,  ±V^T,  ±V^S. 

3.   a*-7x4+Uxi-8  =  0.  4.   x^-lx'+lQx2-  12  =  0. 

5.  x6-10a;4  +  31^-30  =  0.         6.   x6 -  13  a4 +  35  a;2 +  49=0. 

7.   a?-16rf+85a?-150=0.  8.  ^-15a;4+74ar-l20  =  0. 

9.   #6-7a;4  +  7arJ  +  l5  =  0.  10.   x7  -  6  ^  +  3  ar*  +  10  x  =  0. 

11.  x-8-  14^  +  65  a4 -124^  +  84  =  0. 

12.  z8-5;c6-10z4+20arJ  +  24  =  0. 

13.  ar,-9«7+14ari  +  36ar5-72a:  =  0. 

14.  x?  + 11  x-6  +  41  x*  +  61  x2  +  30  =  0. 

15.  z10-4a8-3#6  +  22a;4-4a;2-24  =  0. 

16.  ^-3^-7^-4- 21^+10^-30  =  0. 

17.  x7  -  3  x6  -  10  x5  +  30  x4  -4-  31  ar3  -  93  a2  -  30  a  4-  90  =  0. 

RELATION  OF  ROOTS  TO  COEFFICIENTS 

539.    Prob.     To  find  the  relation  between  the  roots  and  the  coeffi- 
cients of  an  equation. 


324  HIGHER  ALGEBRA 

Solution.  If  a,  b,  c,  are  the  three  roots  of  an  equation,  we 
have  (Art.  520) 

f(x)  =  (x  —  a)  (x  —  b)  (x  —  c)  =  0, 
or  Xs  —  (a  -f  &  4-  c)  #2  4-  («6  4-  ac  4  6c)  cc  —  a6c  =  0. 

If  a,  b,  c,  d,  are  the  four  roots  of  an  equation,  we  have 
f(afy  =  (x  —  a)  (x  —  b)  (x  —  c)(x  —  d)  =  0, 
or     sc4  —  (a  4-  6  4-  c  +  d)  x3  +  (a6  +  ac  +  ad  ■+■  6c  +  6d  -f-  cd)  x2 
—  (abc  4-  a6d  4-  acd  4-  6cd)  ic  +  abed  =  0. 

Generalizing,  we  have  the  following  theorem  : 

The  coefficient  of  xn~l  is  the  sum  of  the  roots  with  their  signs 
changed. 

The  coefficient  of  xn~2  is  the  sum  of  the  products  of  the  roots  taken 
two  and  two. 

The  coefficient  of  xn~3  is  the  sum  of  the  products  of  the  roots,  with 
their  signs  changed,  taken  three  and  three,  and  so  on. 

The  absolute  term  is  the  product  of  all  the  roots  with  their  signs 
changed. 

540.  In  forming  an  equation  from  its  roots  the  student  will  find 
it  safer  and  more  expeditious  to  proceed  as  directed  in  Art.  349. 

Thus,  to  produce  the  equation  whose  roots  are  4,  3,  2,  —  2,  —  1,  we  have 

/(*)  =  (*  -  4)(x  -  3)(x  -  2)(x  +  2)(x  +  1)  =  0. 

The  product  of  the  first  two  factors  is  found  by  the  process  of  Art.  60  to  be 
x2  —  7  x  +  12.     We  then  proceed  as  follows  : 

(_2)       1-7       12 

(2)  _9      26       -24 

(1)  -7        8  28-48 

-6        1  36-20-48 

Hence  the  equation  is 

X5  _  6 x4  +  x3  +  36 x2  -  20x  -  48  =  0. 

If  some  of  the  roots  are  imaginaries  or  quadratic  surds,  and 
hence  occur  in  conjugate  pairs,  the  remainders  should  be  multi- 
plied together  in  pairs,  so  as  to  make  use  of  the  theorem  for  the 
product  of  the  sum  and  difference  of  two  quantities. 


SITUATION  OF  ROOTS  325 

Thus,  to  produce  the  equation  whose  roots  are  2  ±  V3,  1  ±  V—  1,  we  have 

/(x)=(x-2-V3)(x-2+\/3)(x-l  -V^T)(x-  1  +V^T)  =  0, 

or  /(x)  =  (x2-4x  +  l)(x2-2x  +  2)=0. 

The  indicated  multiplication  should  now  be  performed  by  the  process  of 
Art.  56. 

EXAMPLES  CXXX 
Produce  the  equations  whose  roots  are  the  following: 
1.   1,2,3,  -4.  2.   2,5,-1,-3. 

3.   1,  1,  2,  ±V2.  4.   1,  -1,  ±V2,  ±  VS. 

5.   1,  2,  3,  4,  -1,-2.  6.    ±  V2,  ±  V5,  ±  V^I 

7.  if,  -1,  -3,-4.  a  if,  -l,  ±V3. 

9.   2,  -2,  1±V2,  2±V^T.       10.   3±V5,  2±V2,  3±V^3. 

SITUATION  OF   ROOTS 

541.  Theorem.  If  f(x)  has  opposite  signs  tvhen  two  different 
numbers  are  substituted  for  x,  an  odd  number  of  real  roots  lie  be- 
tween the  substituted  numbers;  while  if  fix)  has  the  same  signs  for 
both  numbers,  either  no  real  roots,  or  an  even  number  of  real  roots, 
lie  between  them. 

Dem.  Let  a,  b,  c,  etc.,  in  the  order  of  their  values,  a  being 
least,  be  the  real  roots  of  fix)  =  0,  and  let  <£  (x)  be  the  product 
of  the  remainders  obtained  by  subtracting  the  imaginary  roots, 
if  any,  from  x-,  then  (Art.  520), 

f(x)  =  (*  -a)ix-  b)  ix  -  c)  .-  (<£  ix))  =  0. 

If  any  number  be  substituted  for  x,  all  the  above  factors  con- 
taining roots  less  than  the  substituted  number  will  be  +,  and  all 
containing  roots  greater  than  the  substituted  number  will  be  — . 
If  now  a  second  number  be  substituted  for  x,  only  those  factors 
containing  roots  between  the  substituted  numbers  will  change 
sign.  Hence,  since  <f>  ix)  is  always  +  (Art.  522),  a  change  of  sign 
of  the  product  would  indicate  that  an  odd  number  of  real  roots  lie 


326  HIGUEB  ALGEBRA 

between  the  substituted  numbers ;  while  no  change  of  sign  of  the 
product  would  indicate  that  no  real  roots  or  an  even  number  of 
real  roots  lie  between  the  substituted  numbers. 

Or  the  truth  of  the  theorem  may  be  seen  thus  : 

All  values  <  0  are  —  and  all  values  >  0  are  + .  Now  if  one 
value  of  x  makes /(aj)  — ,  i.e.,  too  small,  and  another  value  of  x 
makes  f(x)+,  i.e.,  too  large,  there  must  be  a  value  of  x  between 
these  two  that  will  make  f(x)  0,  i.e.,  satisfy  the  equation. 

In  passing  from  a  value  too  small  to  a  value  too  large,  i.e., 
from  —  to  +,  or  vice  versa,  f(x)  might  pass  through  0  three,  five, 
or  some  other  odd  number  of  times;  but  if  for  two  different 
values  of  x,f(x)  has  the  same  sign,  it  is  because  f(x)  has  either 
not  passed  through  0  at  all,  or  passed  through  an  even  number  of 
times,  leaving  it  on  the  same  side  of  0. 

542.  Cor.  Whenever  the  substitution  of  two  consecutive  numbers 
gives  opposite  signs  to  f(x),  the  numerically  smaller  of  the  two  num- 
bers is  the  integral  part  of  the  root  that  lies  between  them. 

543.  Sch.  1.  While  we  may  find  the  situation  of  negative  roots 
by  evaluating  f(x)  for  negative  numbers,  it  is  a  little  more  con- 
venient to  evaluate  f(—x)  for  positive  numbers  (Art.  526).  Be- 
sides, as  will  be  seen  later,  in  approximating  beyond  the  integral 
part  the  value  of  a  negative  root,  it  is  necessary  to  employ /(—  x). 

544.  Sch.  2.  By  taking  the  value  of  x  large  enough  the  first 
term  off(x)  may  be  made  larger  than  any  or  all  of  the  other  terms. 
Hence,  if  a  series  of  minus  signs  result  from  the  substitution 
of  larger  and  larger  numbers,  the  situation  of  a  root  will  be  found 
by  taking  still  larger  numbers,  as  the  function  must  eventually 
become  plus,  the  sign  of  the  first  term. 

EXAMPLES   CXXXI 

Find  the  situation  of  all  the  real  roots  of  the  following : 

1.   a4-2arJ-13a2  +  22a  +  22  =  0. 

Solution.  Since  f(x)  has  but  two  variations,  the  equation  cannot  have 
more  than  two  positive  roots  (Art.  529).     Substituting  for  x,  by  Art.  513, 


8ITUATION   OF  ROOTS 


327 


the  values  0,  1,  2,  3,  etc.,  we  have  for  /(as)  the  signs  given  in  the  margin, 
showing  that  the  positive  roots  lie  between  2  and  8,  3 
and  4. 

Changing  the  signs  of  the  alternate  terms  of  f(x), 
we  have 


/(-  x)  =  x*  +  2  x3  -  13  &  -  22  x  +  22  =  0. 


/(*) 

X 

/(-*) 

+ 

0 

+ 

-f 

1 

— 

+ 

2 

— 

- 

3 

- 

+ 

4 

+ 

The  positive  roots  of  this  equation  are  seen  to  lie 
between  0  and  1,  3  and  4;   therefore  (Art.  526)  the  negative  roots  of  the 
original  equation  lie  between  0  and  —1,-3  and  —  4. 

2.   «3-3x2-4ic  +  ll  =  0.         3.  x3-3x2-10x-\-5  =  0. 
4.  x*  +  Q>tf  +  2x-  1  =  0.  5.   x*-  Gx2  +  2x  +  2  =  0. 

6.   «4-4^-3xH-23  =  0. 

Sue  Since  f(x)  has  but  two  variations,  the  equation  cannot  have  more 
than  two  positive  roots  (Art.  529). 

Since  all  the  signs  of  f{—x)  are  +,  /(—  x)  =  0  can  have  no  positive 
roots  (Art.  529);  hence  f(x)=  0  can  have  no  negative  roots.  It  follows  that 
two  of  the  four  roots  are  imaginary. 

7.  ar3-2«2-10  =  0.  8.   ar3-3a;-l=0. 

9.  x*-2x2-6x-2  =  0.  10.   x3-5x2-4:  =  0. 

11.   ar3  +  ar>- 10  =  0. 

Solution.    The  position  of  the  one  positive  root  is  easily  found. 
Changing  the  signs  of  the  terms  containing  the  even  powers  (Art.  526), 
we  have 

/(-Z)=£3-X2+  10=0. 

This  function  can  become  0  only  by  having  the  single  negative  term  numeri- 
cally equal  to  the  sum  of  the  two  positive  terms.  Now,  if  x  <  1,  the  10  alone, 
without  the  x3,  is  more  than  enough  to  counterbalance  the  negative  term  ; 
while  if  x>  1,  the  x3  alone  without  the  10,  is  more  than  enough  to  counter- 
balance the  negative  term. 

Hence  this  equation  can  have  no  positive  roots,  and  the  original  equation 
no  negative  roots. 

12.  xs+xi+x-50=0.  13.   ic5 4-^+^-20=0. 

14.  8^-36^+46^-15=0.    15.   z4-10ar3+31  ^-30^+6=0. 
16.  »4-6arJ+4x2+18a;-21=0. 


328  HIGH  Ell   ALGEBRA 

Solution.  The  one  variation  of  /(—  x)  shows  that  the  equation  cannot 
have  more  than  one  negative  root.     A  negative  root  is  readily  located  between 

-  1  and  -  2. 

x  I  f  Cx~) 
Evaluating  f(x)  for  positive  numbers,  the  results  are  as  in  the     _  J  v  ' 

margin,-  locating  one  root  between  4  and  5.     As  all  numbers  >5  0—21 

give  positive  results,  the  other  two  roots  must  either  both  beimagi-  1—4 

nary,  or  both  be  situated  between  two  consecutive  numbers.     It  2    —    1 

is  observed  that  both  1  and  2  very  nearly  satisfy  the  equation  (caus-  3—12 

ing  f(x)  to  differ  little  from  0),  suggesting  that  two  roots  may  lie  4—13 

between  these  numbers.     Evaluating  for  1.1,  1.2,  1.3,  etc.,  we  find  5    + 

changes  of  signs  between  1.5  and  1.6,  and  between  1.7  and  1.8,  thus  locating 

the  other  two  roots. 

17.   2ars-3a2-6#  +  9  =  0. 

Sug.  Proceeding  as  in  the  preceding  example,  it  is  rendered  probable 
that  there  are  two  roots  between  1  and  2.  Evaluating  for  the  tenths,  1.5  is 
found  to  be  a  root.  The  best  way  to  find  the  remaining  two  roots  is  to 
depress  the  equation  and  solve  the  resulting  quadratic. 

Or  this  equation  may  be  transformed  into  one  of  the  normal  form,  in  which 
form  it  has  one  integral  root. 

18.  ^-2^-16^  +  24^  +  48  =  0. 

19.  x*-4:Xs-x2  +  12x-6  =  0. 

20.  x5-6x4  +  36x2-25x-30  =  0. 

Sug.  Remove  the  factor  containing  the  integral  root  and  use  the  depressed 
equation. 

21.  x?-5x4  +  3x3  +  16x2-26x  +  ll  =  0. 

22.  x5-2x3-x2-3x-l  =  0. 

23.  .?4-l0ar3  +  33a2-40cc  +  i4  =  0. 

MULTIPLE    ROOTS 

545.  Theorem.  If  an  equation  f(x)  =  0  has  equal  roots,  they  are 
the  roots,  and  the  only  roots,  of  the  equation  formed  by  placing  equal 
to  0  the  highest  common  divisor  of  f(x)  and  its  first  differential 
coefficient. 

Dem.  Let  a  be  one  of  the  m  equal  roots  of  f(x)  =  0.  Then 
(x  —  a)m  is  a  factor  of  f(x).  If  we  represent  the  product  of  the 
other  factors  by  <f>  (x),  we  have 

/(*)  =  (*- «r  <H*)  =  o.  a) 


MULTIPLE  ROOTS  329 

Representing  the  first  differential  coefficients  of  f(x)  and  <ft  (x) 
by  f'(x)  and  <f>'(x)  respectively,  we  have  (Arts.  415  and  420) 

f'(x)  =  m  (x  -  a)—1  <f>  (x)  +  (x-  a)m  <f>'(x)  =  °-  (2) 

Comparing  (2)  with  (1),  we  see  that  (x  —  a)**--1  is  the  h.  c.  d.  of 
f(x)  and  f'(x),  and  that  the  equation, 

(x  -  a)m~l  =  0, 

has  m  —  1  roots  equal  to  a,  and  has  no  other  roots. 

Similarly,  if  a  is  one  of  the  m  equal  roots  and  b  one  of  the  n 
equal  roots  of  f(x)  =  0,  we  shall  have 

(x  -  a)"1-1  (x  -  b)^1  =  0, 

which  has  m  —  1  roots  equal  to  a  and  n  —  1  roots  equal  to  6,  and 
has  no  other  roots. 

546.  The  work  of  finding  the  h.  c.  d.  of  f(x)  and  f'(x)  is  so  great 
that  the  process  should  be  resorted  to  only  when  shorter  processes 
fail.     It  may  be  avoided  in  the  following  cases : 

1.  Multiple  integral  roots  are  found  by  the  same  operation 
that  gives  other  integral  roots  (Art.  537). 

2.  Multiple  monomial  imaginary  and  quadratic  surd  roots  in 
equations  containing  only  even  powers  or  only  odd  powers  are 
found  as  in  Art.  538. 

3.  Multiple  roots  in  an  equation  in  which  f(x)  is  a  perfect 
square  are  found  from  the  equation  obtained  by  extracting  the 
square  root. 

4.  Finally,  in  obtaining  multiple  roots,  it  is  never  necessary  to 
find  the  h.  c.  d.  of  f(x)  and  f'(x)  for  equations  below  the  6th 
degree,  as  seen  from  the  following  considerations  : 

(a)  If  a  cubic  has  a  multiple  root,  we  must  have  either 

(x  -  af  =  0  or  (x  -  a)2  (x-b)  =  0. 

In  both  forms,  to  give  rational  coefficients,  the  roots  must  be 
commensurable. 

(b)  If  a  biquadratic  has  multiple  roots  we  must  have  one  of 
the  forms 

(x  -  a)4  =  0,  (x  -  af  (x  -b)=0,  (x  -  a)2  (x  -b)(x-c)=  0, 

or  (x  —  a)2  (x  —  b)2  =  0. 


330  HIGHER   ALGEBRA 

In  the  first  three  forms,  to  give  rational  coefficients,  the  multiple 
roots  must  be  commensurable.  In  the  fourth  form  a  and  b  can  be 
incommensurable  if   they  are  numerically  equal  with  opposite 

signs,  giving 

(x  -  a)2  (x  +  a)2  =  (x2  -  a2)2  =  0. 

In  this  case  f(x)  is  a  perfect  square. 

(c)  If  a  quintic  has  multiple  roots,  we  must  have  one  of  the 
forms 

(x  -  a)5  =  0,  (x  -  a)4  (x  -  b)  =  0,  (x  -  a)3  (x  -b)(x-c)  =  0, 

(x  -  a)8  (x  -  bf  =  0,   (x  -  a)2  (x  -b)(;x-  c)(x  -  0)  =  0, 

or  (x-a)2  (x-b)2(x-c)  =  0. 

In  all  but  the  last  of  these  forms,  to  give  rational  coefficients, 
the  multiple  roots  must  be  commensurable.  In  the  last  form 
a  and  b  can  be  incommensurable  if  they  are  numerically  equal 
with  opposite  signs,  giving 

(x  -a)2  (x  +  a)9  (x-c)  =  (x2-  a2)2  (x-c)=  0. 

In  this  case  the  commensurable  root  c  may  be  removed  (Art. 
515),  leaving  a  perfect  square. 

EXAMPLES  CXXXII 

Find  the  multiple  roots  of  the  following : 

1.   x«-±x6-x*  +  16x3-5x2-12x-3  =  0. 

Solution.  Since  the  equation  is  of  even  degree  and  the  absolute  term 
— ,  it  has  at  least  two  real  roots,  and  these  have  opposite  signs  (Art.  525). 
As  indicated  in  the  margin,  there  are  no  changes  of  sign  except  between 
1  and  2,  and  between  —  1  and  —  2. 


/GO 


o 
-         l 

+  2 

+  3 


/(-*) 


+ 


Now,  if  any  of  the  remaining  roots  are  real,  they 
must  be  situated  either  between  these  same  numbers  q 
(a  change  of  sign  indicates  a  passage  through  an  odd  ^ 
number  of  roots),  or  in  pairs  between  other  con-  % 
secutive  numbers.  We  therefore  apply  the  test  for  g 
equal  roots. 

Obtaining  the  first  differential  coefficient,  we  have 

f'(x)  -  6  xb  -  20x4  -  4  xs  +  48  x2  -  10  X  -  12. 

Rejecting  the  factor  2,  finding  the  h.  c.  d.  of  f(x)  and/(x),  and  placing  it 
equal  toO,  we  have  x2_2x_1  =  0, 

or  x  =  1  ±  V2. 


SOLUTION  FOR  INCOMMENSURABLE  ROOTS  331 

Therefore  1  +  v/2  and  1  —  V2  are  each  double  roots. 

Again,  since  x2  —  2  x  —  1  is  the  h.  c.  d.  of  /(x)  and  f(x),  (x2  -  2  x  —  l)2, 
orx*-4x3  +  2x2  +  4x  +  l,  is  a  factor  of /(x)  (Art.  545).  The  other  factor 
is  found  by  division  to  be  x2  —  3,  giving  for  the  other  roots  x  =  ±  V3.  Hence 
the  six  roots  are  1  ±  V2,  1  ±  V2,   ±  VS. 

2.  a;4-8ar3  +  18a;2-8a;-f- 1  =  0. 

3.  z4-12arJ  +  50#2-84a;  +  49  =  0. 

4.  ar}-7a;4  +  8ar}  +  28arJ-22a;-48=0. 

5.  x6-7x4-}-Wx2-12  =  0. 

6.  a;6-2ar5-6a;4  +  8ar}  +  12aro-8a;-8  =  0. 
1.  ^-12^  +  45^-50^  =  0. 

8.  x«-8x5  +  12x4  +  24:X3-27x2-lGx-l=0. 

9.  a?-9a^5  +  30a;4-44ar2-f-24  =  0. 

10.   x8 -  8x7  +  16a;6  +  16a,-5  -  56a;4  -  32ar*  +  64a;2  +  64a;  +  16=  0. 

SOLUTION  FOR   INCOMMENSURABLE   ROOTS 
Horner's  Method 

547.  Having  explained  how  to  find  the  situation  and,  conse- 
quently, the  initial  figures  of  the  real  roots  of  an  equation,  we 
now  proceed  to  exhibit  a  process  for  finding  the  remaining  parts 
of  such  roots,  exactly  if  commensurable,  and  to  any  required 
degree  of  approximation  if  incommensurable.  This  process  is 
called  Horner's  Method,  having  been  published  by  W.  G.  Horner, 
of  Bath,  England,  in  1819.  The  method  is  based  on  the  next  two 
theorems. 

548.  Theorem.  If  the  first  member  of  an  equation  in  the  form 
f(x)  =  0  be  divided  by  x  —  a,  then  the  integral  part  of  the  quotient  be 
again  divided  by  x—  a,  and  so  on,  the  successive  remainders  ivill  be, 
in  inverse  order,  the  coefficients  of  an  equation  whose  roots  are  less 
by  a  than  those  of  the  given  equation. 

Dem.     Let  the  given  equation  be 

Asf  +  Bxn~l  +  Cxn~2  H Jx-  +  Kx  +  L  =  0.  (1) 

Let  x  =  xx  +  a.     Then  by  substitution 

A(x,  +  a)n  +  B(xx  +  a)n~l  +  C(xx  +  a)n~2+  •  •  •  J(x,  +  af 
-f-  K(xx  +  a)  +  L  =  0. 


332  HIGHER  ALGEBRA 

If  we  should  perform  the  indicated  operations  and  arrange 
with  reference  to  xl}  there  would  result  an  equation  of  the  form 

Ax?  +  B&f-1  +  Cxx?~2  H J^2  +  A>,  +  A  =  0.         (2) 

The  unknown  quantity  in  (1),  the  given  equation,  is  x,  while 
in  (2),  the  transformed  equation,  it  is  xx,  which  was  taken  less  by 
a  than  x.  Hence,  the  roots  (which  are  the  values  of  the  unknown 
quantity)  of  (2)  are  less  by  a  than  those  of  (1). 

In  this  transformation  all  the  coefficients  except  the  first  have 
been  changed.  Restoring  in  (2)  the  value  of  xx,  which  is  x  —  a, 
we  have 

A(x  -  a)n  +  Bx(x  -  a)^1  +  Ct(x  -  a)n~2  +  ....  J%(x  -  af 

+  K1(x-a)  +  Ll  =  0*  (3) 

which  must  be  identical  with  (1),  though  in  different  form.  It  is 
seen  that  if  we  divide  the  first  member  of  (3)  by  x  —  a,  then 
divide  the  quotient  by  x  —  a,  and  so  on,  the  successive  remainders 
will  be  Llf  K\,  J\,  •"  Cp  Bl}  A,  the  coefficients,  in  inverse  order, 
of  (2),  which  is  an  equation  whose  roots  are  less  by  a  than  those 
of  the  given  equation. 

EXAMPLES  CXXXIII 

1.   Find  the  equation  whose  roots  are   each  less  by  2   than 

those  of 

x*  -  1  x>  +  llx2  +  1  x  -12  =0. 

*  The  coefficients  L\,  K\,  J\,  •••  C\,  B\,  A  are,  respectively,  the  function, 

..  „  .  7  .  ..  M  derivative  3d  derivative  nth  derivative  ..,  m 
its   1st  derivative, ,  — ,   •••  : ,  with    x 

replaced  by  a.  •—  •— 

Substituting  Xi  +  a  for  x  in  f(x)  =  0  and  developing  by  Taylor's  For- 
mula, using  the  form  in  Art.  452,  we  have 

/(*)  =f(xl  +  d)  =/(«)  +/'(«)*i+/"(«)  ^  +/'"(«)  f£  +/iv(«)^  +  etc.  =  0. 

2  t?  I4 

Hence  Ll=f(a),  Kx  =f>(a),  Jt=C&K  '"  &=£=%  A  =£&>. 

2  [n  —  1  \n 

If  X\  is  a  small  fraction,  for  an  approximate  value  the  higher  powers  may 
be  neglected,  and  we  have 

f'(a)x\  -\-f(a)=  0,  whence  x\  —  —   .^r\'  aPProx'mately-     [See  Art.  549.] 


SOLUTION  FOR  INCOMMENSURABLE  ROOTS  333 

Solution.  By  the  theorem  the  successive  remainders  obtained  by  dividing 
by  x  —  2,  then  dividing  the  quotient  by  x  —  2,  and  so  on,  will  be,  in  inverse 
order,  the  coefficients  sought.  These  remainders  are  obtained  with  great 
facility  by  synthetic  division,  as  follows  : 

x4  -  7  x3  +  11  x2  +  7  x  -  12  |_2 

-  5  1  9  6     1st  rem. 

-  3        —  5      —  1    2d  rem. 

-  1       -  7    3d  rem. 
1    4th  rem. 

1     5th  rem. 

Hence  the  equation  whose  roots  are  each  less  by  2  than  those  of  the  given 

equation  is 

Xi*  +  Xi3  -  7  xi2  -  xi  +  6  =  0, 

or,  omitting  subscripts,       x4  +  x3  —  7x2  —  x  +  6  =  0. 

Let  the  student  find  by  Art.  537  the  roots  of  each  equation,  and  see 
whether  those  of  the  second  are  less  by  2  than  the  corresponding  ones  of  the 
first. 

2.  Find  the  equation  whose  roots  are  less  by  3.14  than  those  of 
2xA-5xi-3xi  +  4:X-23  =  0. 

Solution.  We  may  diminish  the  roots  first  by  3,  then  by  .1  more,  and 
then  by  .04  more,  without  writing  the  intermediate  equations,  as  follows  : 

2  x*  -  5  x3  -  3  x2  +  4  x  -  23  [_3 

4  -11(1)  [A 

67(1)  7.3192 

6.192  -  3.0808(2)  [M 

73.192  3.28970432 


-5x3 

-  3  x2 

1 

0 

7 

21 

13 

60(i) 

19(1) 

1.92 

.2 

61.92 

19.2 

1.94 

.2 

63.86 

19.4 

1.96 

.2 

65.82(2) 

19.6 

.7952 

.2 

66.6152 

19.8(2) 

.7984 

.08 

67.4136 

19.88 

.8016 

.08 

68.2152(3) 

6.386  -  .39109568(3) 

79.578(2) 
2.664608 


82.242608 
2.696544 
84.939152(3) 


19.96 

.08 
20.04 

.08 
20.12(8) 


334  HIGHER   ALGEBRA 

The  numbers  marked  (1),  together  with  the  first  coefficient,  which  re- 
mains unchanged,  are  the  coefficients  of  the  equation  whose  roots  are  less 
by  3  than  those  of  the  given  equation. 

The  numbers  marked  (2) ,  together  with  the  first  coefficient,  are  the  coeffi- 
cients of  the  equation  whose  roots  are  less  by  .  1  than  those  of  the  equation 
whose  coefficients  are  marked  (1),  and,  consequently,  less  by  3.1  than  those 
of  the  given  equation. 

The  numbers  marked  (3),  together  with  the  first  coefficient,  are  the 
coefficients  of  the  equation  whose  roots  are  less  by  .04  than  those  of  the 
equation  whose  coefficients  are  marked  (2),  less  by  .14  than  those  of  the  equa- 
tion whose  coefficients  are  marked  (1),  and  less  by  3.14  than  those  of  the 
given  equation.  Hence  the  equation  whose  roots  are  less  by  3.14  than  those 
of  the  given  equation  is 

2  x4  +  20.12  x*  +  68.2152  x2  +  84.939152  x  -  .39109568  «c  0. 

3.  Find  the  equation  whose  roots  are  less  by  3.213  than  those 
of  x3  -f- 11  x2  -  102  x  +  181  =  0. 

4.  Find  the  equation  whose  roots  are  less  by  2.85  than  those 
of  a4-12x2  +  12  a? -.3  =  0. 

549.  Theorem.  When  a  root  of  an  equation  is  a  small  fraction, 
it  is  approximately  equal  to  the  absolute  term  divided  by  the  coefficient 
of  the  first  power  of  x. 

Dem.     Let  x1}  a  small  fraction,  be  a  root  of  the  equation 

Axn  +  Bxn~l  +  Cxn~2  +.--Kx+L  =  0. 
Substituting,  we  have 

As?  +  Bxf-1  +  Cx?-2  +  ••  •  Kxx  +L  =  0. 

Now  since  x1  is  a  small  fraction,  the  terms  with  powers  above 
the  first  are  so  small  that  the  equation  will  be  little  affected  by 
neglecting  them  and  retaining  only  Kxt  +  L  =  0,  which  gives 

xx  = ,  approximately. 

K 

Illustration.     Consider  the  equation 

9X3_X2  +  9X_  i  _o. 

When  x  =  0,  f(x)  =  -1,  and  when  x  =  1,  f{x)  =  +  16. 
Hence,  there  is  a  real  root  between  0  and  1,  and  it  is  much  nearer  0  than 
1,  i.e.,  it  is  a  small  fraction.    Then  x3  and  x2  are  much  smaller  than  x,  and 


SOLUTION   FOR  INCOMMENSURABLE  ROOTS  335 

the  equation  will  be  little  affected  by  neglecting  the  terras  containing  them, 
giving  9  x  —  1  =  0,  whence  X  a  £  =  .11,  approximately.  Indeed,  in  this  case, 
the  quotient,  $,  of  the  absolute  term  divided  by  the  coefficient  of  the  1st 
power  of  x  is  the  exact  root. 

550.  The  principle  of  Horner's  Method  of  finding  a  root  of  an 
equation,  exactly  if  commensurable,  approximately  if  incommen- 
surable, is  this :  Suppose  the  integral  part  of  the  root  to  have 
been  found  by  Art.  542.  If  by  Art.  548  we  find  the  equation 
whose  roots  are  less  by  this  integral  part  than  those  of  the 
given  equation,  the  corresponding  root  of  this  transformed  equa- 
tion will  be  the  decimal  part  of  the  root  of  the  given  equation. 
Now,  by  Art.  549,  this  part  is  approximately  equal  to  the  abso- 
lute term  of  this  transformed  equation  divided  by  the  coefficient 
of  the  1st  power  of  x  in  the  same  equation  (i.e.;  the  first  remain- 
der divided  by  the  second  remainder  obtained  in  successively 
dividing  f(x)  by  x  minus  the  integral  part  of  the  root).  Using 
only  the  first  decimal  figure  of  the  quotient,  we  may  find  the- 
equation  whose  roots  are  less  by  this  decimal  figure  than  those  of 
the  first  transformed  equation,  and,  as  before,  obtain  another 
figure  of  the  root  by  dividing  the  absolute  term  of  this  second 
transformed  equation  by  the  coefficient  of  the  1st  power  of  x  in 
the  same  equation,  and  so  on. 

551.  Reverting  to  example  2,  page  333,  we  may  see  an  application 
of  the  method.  By  the  principle  of  Art.  541,  it  is  found  that  a  real 
root  lies  between  3  and  4,  i.e.,  one  root  is  3  +  a  decimal  fraction. 

The  equation  whose  roots  are  less  by  3  has  for  its  correspond- 
ing root  this  decimal  fraction,  and  this  fraction,  by  Art.  549,  is 
approximately  the  first  remainder  divided  by  the  second,  i.e., 
11  -5-  67  =  .1,  approximately,  using  only  the  first  decimal  figure. 

The  equation  whose  roots  are  less  by  .1  than  those  of  the  last 
has  for  its  corresponding  root  the  remaining  figures  of  the  deci- 
mal fraction,  and  this  is  approximately  the  new  first  remainder 
divided  by  the  new  second  remainder,  i.e.,  3.6808  -j-  79.578  (it  is 
sufficient  to  use  3.68  -r-  79)  =  .04,  approximately. 

For  the  third  and  probably  the  fourth  figures  of  the  decimal 
part  of  the  root  we  have  .39109568  --  84.939152  =  .0046.  Hence 
the  root  to  four  decimal  places  is  3.1446. 


336  HIGHER  ALGEBRA 

For  a  farther  application  of  the  method,  let  it  be  required  to 
find  to  four  decimal  places  the  positive  root  of 

a? +  2^ -9a; -22  =  0.  x   f(z) 

As  indicated  in  the  margin,  the  one  positive  root  lies  between  0 
3  and  4,  i.e.,  it  is  3  +  a  decimal  fraction.  We  must,  therefore,  1 
diminish  the  roots  by  3,  and  for  a  second  figure  of  the  root,  divide  2 
the  first  remainder  by  the  second,  and  so  on,  the  work  being  as  3 
follows :  4 

x3  +  2x2  -  9 x  -  22 J3.1273 

5  6-    4(i) 

8  30(i)  3.111 

11(D  l.U  -  .889(2) 

.1  31.11  .649128 

11.1  1.12  -  .239872(3) 
.1                         32.23(2) 

11.2  .2264 
.1                          32.4564 

11.3(2)  .2268                           1st  figure  =  3, 

.02  32.6832(3)                       2d  figure  =  4  +  30  =  .1, 

11.32  3d  figure  =  .889  +  32.23  =  .02, 

.02  4th  and  5th  figures  =  .239872  4-  32.6832  =  .0073 


11.34 


552.    Carefully  note  the  following  observations : 

1.  As  we  find  a  root  by  starting  with  a  value  too  small  and 
increasing  it  by  annexing  figure  after  figure,  a  change  of  sign  of 
the  absolute  term  in  any  of  the  transformed  equations,  i.e.,  the  first 
remainder  of  any  of  the  successive  sets  of  divisions,  would  indicate 
that  we  had  passed  beyond  the  value  of  the  root  (Art.  541),  and  that 
the  last  figure  must  be  diminished.  The  figure  to  be  adopted  in 
any  case  is  the  largest  number  which,  in  the  process  of  diminishing 
the  roots,  will  not  make  the  sign  of  the  absolute  term  different  from 
that  of  the  first  transformed  equation.  If  the  absolute  term  of  the 
first  transformed  equation  is  different  from  that  of  the  given  equa- 
tion, it  simply  indicates  that  there  is  another  root  between  0  and 
the  one  we  have  started  to  find. 

2.  As  the  successive  figures  of  the  root  are  found  by  neglecting 
all  but  the  last  two  terms  of  an  equation  whose  roots  are  less  by 
the  part  of  the  root  already  found  than  those  of  the  given  equa- 


SOLUTION  FOR  INCOMMENSURABLE  ROOTS  337 

tion  (Art.  549),  the  quotient  of  the  absolute  term  of  a  transformed 
equation  by  the  coefficient  of  the  1st  power  of  x  in  the  same  equa- 
tion is  liable  to  be  too  large,  though  seldom  so  beyond  the  second 
figure  of  the  root.  To  avoid  this  for  the  second  figure  of  the  root, 
we  may  solve,  approximately,  the  quadratic  obtained  by  neglect- 
ing, not  all  but  two,  but  all  but  three,  of  the  terms  of  the  first 
transformed  equation.  This  is  also  necessary  if,  in  any  case,  the 
next  to  the  last  term  is  0. 

3.  Since  x1  =  —  -—,  approximately  (Art.  549),  in  which  L  is  the 

first  and  K  the  second  remainder,  K  and  L  must  have  opposite 
signs,  for  otherwise  the  quotient  would  be  the  amount  to  be  sub- 
tracted from,  instead  of  added  to,  the  part  of  the  root  already 
found.  If  after  the  first  transformation  the  first  and  second 
remainders  have  like  signs,  the  next  figure  of  the  root  cannot  be 
found  by  division,  but  must  be  found  by  the  same  kind  of  trial 
which  gives  the  first  figure. 

4.  If  two  or  more  roots  have  the  same  initial  part,  the  next 
figure  of  each  must  be  found  by  trial,  after  which  the  process  is 
the  same  as  for  other  cases. 

5.  Ordinarily  the  fourth  decimal  figure  will  be  correctly  given 
by  writing  two  figures  of  the  third  quotient. 

6.  The  negative  roots  of  an  equation  are  found  by  finding  the 
positive  roots  of  the  equation  obtained  by  changing  the  signs  of 
the  odd  or  the  even  powers  (Art.  526). 

7.  When  all  but  one  of  the  roots  have  been  found,  the  remain- 
ing one  becomes  known  by  adding  the  sum  of  the  known  roots  to 
the  coefficient  of  the  second  term  (the  coefficient  of  #n_1)  of  the 
given  equation  and  changing  the  sign  of  the  result  (Art.  539). 


EXAMPLES   CXXXIV 

Find  to  three  or  four  decimal  places  the 
real  roots  of  the  following : 

1.  x3-4:X2-Gx  +  8  =  0. 

Solution.     As  indicated  in  the  margin,  the  posi- 
tive roots  are  between  0  and  1,  4  and  5,  while  the 
negative  root  is  between  —  1  and  -  2. 
downey's  alg. — 22 


X 

m 

X 

0 

+ 

0 

1 

— 

1 

2 

— 

2 

3 

— 

4 

— 

5 

+ 

/(-*) 


338  Mgher  Algebra 

To  find  the  negative  root  the  operation  is  as  follows  : 

-8        [1.8004 

-9(1) 
8.992 


xs  +  4  x2 
5 

-6x 
1 

6 

7(1) 

5(1) 
6.24 

•008(2) 


_  11.24 

7.8  6.88  Since    9-f  5   cannot   give    the 


18.12(3) 


8.6 


first  decimal  figure,  we  use 

7x2  +  5:s-9  =  0,  (Obs.  3) 


.g  whence,  x2  +  f  x  =  f, 

9.4(3)  and  g~     *  ±*L*SS.8. 

The  2d  quotient  =  .008  +  18.12  =  .0004. 

Let  the  student  show  that  the  root  between  4  and  5  is  4.8922.   y/^ 
The  sum  of  these  two  roots  and  the  second  coefficient  is  —  .9082.     Hence 
the  third  root  is  .9082  (Obs.  7). 

2.    8  a8 -17  a2 -16  a; +  34  =  0. 

Solution.  By  trial  the  roots  are  seen  to  lie  between  1  and  2,  2  and  3, 

—  1  and  —  2.  The  operation  of  finding  the  largest  root  is  as  follows  : 
8x3-17x2  -16  a;  +  34  |  2.125 

-    1  -18  -    2(i) 

15  12(i)  1.518 

31(i)  JU8  -      .482(2) 

.8  15.18  .382224 

31.8  3.26  -      .099776(3) 

.8  18.44(2)  .099776 

32.6  .6712  000000 


.8  19.1112 

33.4(2)  .6744 


1st  figure  =  2, 


.16  19.7856(3) 

3^56  .1696  2d  figure  =  2^  12  =  .1, 

.16  19.9552  3d  figure  =  .482  +  18.44  =  .02, 

33. 72  4th  figure  =  .099776  -  19.7856  =  .005. 

.16  As  the  absolute  term   reduces  to  0,  the  exact  root 

33.88(3)  has  been  found. 

.04  The  other  two  roots  might  also  be  found  by  Horner's 

33.92  Method  ;  but  since  this  root  is  commensurable,  and  its 

exact  value  has  been  found,  a  much  better  way  is  to 
depress  the  equation  and  solve  the  resulting  quadratic,  as  follows  : 

8x3-17x2-16x  +  34  LgJ 

0-16  0 


SOLUTION  FOR  INCOMMENSURABLE  ROOTS  339 

Hence  8  x2  -  16  =  0,  whence  x2  =s  2,  and  x  =  ±  V2. 

3.  a*-6a?  +  5x  +  ll  =  0.  4.  ar3  +  10ar>  + 5a; -260  =  0. 

5.  8a?-65x*+ 140 z-33=0.  6.  ar5  +  3x2  +  4a;  +  5  =  0. 

7.  ar»  +  a;  =  1000.  8.  4ar*- 9ar*-52a;  + 117  =  0. 

9.  ic3-3x2-3x  =  -18.  10.  «3-3»2-4^  +  13  =  0. 

11.  x4  +  4ar3-5arJ-18a;  =  22.  12.  x4 -6x* +  3x> +  30x  =  50. 

13.  ^+4^-4^-11x4-4=0.  14.  2a,-4+5ar3+4ar2+3a;=8002. 

15.  ar5-4x-2000  =  0.  16.  ar5  -  4  a;4 +- 7  ar*  -  863  =  0. 

17.  The  radius  of  a  sphere  is  9  inches,  and  the  volume  of  a 
segment  of  one  base  cut  from  it  is  one  fourth  of  that  of  the 
sphere.  Find  its  altitude,  the  volume  of  a  spherical  segment  of 
one  base  in  terms  of  its  altitude  and  the  radius  of  the  sphere 
being  ir(rh2--^\ 

125  7T 

18.  The  weight  of  a  sphere  1  foot  in  diameter  is  — - —  pounds. 
When  floated  in  a  full  vessel  of  water  weighing  62.5  pounds  per 

cubic  foot,  it  causes  an  overflow  of  -^-  cubic  feet  of  water.     Find 

16 

the  depth  to  which  it  sinks,  the  law  being  that  the  weight  of 
the  water  displaced  equals  the  weight  of  the  floating  body. 

19.  A  rectangular  inner  court,  100  feet  long  and  50  feet  wide, 
has  its  principal  openings  at  diagonally  opposite  corners.  Between 
these  openings  matting  7  feet  wide  and  with  square  ends,  the 
corners  just  touching  the  walls,  is  laid  (i.e.,  the  matting  is  a 
rectangle  inscribed  within  a  rectangle).  Find  the  length  of  the 
matting. 

553.  The  extraction  of  the  nth  root  of  any  number,  as  a,  is  the 
same  as  finding  the  positive,  real  root  of  the  equation, 

xn  +  Oaj""1  +  0xn-2  -\ a  =  0. 

Hence  Horner's  Method  will  give  to  any  required  degree  of 
accuracy  any  root  of  a  given  number.  By  the  principle  of  Art. 
471,  the  use  of  a  table  of  logarithms  will  give  the  result  with  far 
less  labor  than  will  either  the  elementary  method  or  Horner's 
Method. 


340  HIGHER   ALGEBRA 

EXAMPLES   CXXXV 

Perform  by  Horner's  Method  the  following  indicated  opera- 
tions : 

1-   -\/29791.  2.    V70444997. 

3.    ^5\6847.  4.    4/B. 

Sturm's  Theorem  and  Method 

554.  The  change  of  sign  of  f(x)  as  we  pass  from  one  value  of 
x  to  another  usually  reveals  with  little  difficulty  the  situation  of 
the  real  roots  of  an  equation  (Art.  541).  In  infrequent  cases, 
however,  there  are  difficulties,  the  nature  of  them  being  this: 
After  locating,  by  Art.  541,  all  the  real  roots  which  the  substitu- 
tion of  integral  numbers  will  reveal,  we  are  sometimes  uncertain 
whether,  1st,  the  remaining  roots  are  imaginary,  2d,  three  or 
higher  odd  number  of  real  roots  lie  between  two  consecutive 
numbers  that  cause /(a?)  to  change  sign,  or,  3d,  two  or  higher  even 
number  of  real  roots  lie  between  two  consecutive  numbers  that 
do  not  cause  f(x)  to  change  sign.  In  the  latter  case  the  method 
of  ex.  16,  p.  327,  usually  removes  the  uncertainty  and  locates  the 
roots  if  real ;  but  in  exceptional  cases  the  uncertainty  remains  or 
is  removed  only  after  much  labor.  In  1829  Jacques  Charles 
Franqois  Sturm  (1803-1855),  a  Swiss  mathematician,  who  after- 
ward became  a  member  of  the  French  Academy,  Professor  of 
Mathematics  in  the  Polytechnic  School  and  Professor  of  Mechan- 
ics in  the  Faculte  des  Sciences  in  Paris,  discovered  a  theorem  by 
means  of  which,  in  all  cases,  the  number  and  situation  of  the  real 
roots  of  a  numerical  equation  may  be  found.  As  its  application 
is  laborious,  it  is  used  only  as  a  last  resort  when  the  method  of 
Art.  541  fails  or  is  not  readily  applicable. 

555.  If  f(x)  and  its  first  derivative  be  treated  as  in  the  process 
of  finding  the  h.  c.  d  ,  except  that  each  remainder,  before  being  used 
as  the  next  divisor,  have  its  signs  changed  and  that  no  negative 
factors  be  introduced  or  rejected,  these  successive  remainders 
with  their  signs  changed  constitute  what  are  called  Sturm's  Func- 
tions, or  the  Sturmian  Functions. 


SOLUTION  FOR  INCOMMENSURABLE  ROOTS  341 

Thus,  let  f(x)  =  x3  -  4  x'2  -  x  +  4  =  0. 

The  first  derivative  is  3  x*  -  8  x  -  1.  Dividing  x3-4z2-j;  +  4  by 
3  x2  —  8  a;  —  1,  first  multiplying  the  former  by  3  to  avoid  fractions,  as  in  the 
process  of  finding  the  h.  c.  d.  (Art.  Ill),  the  first  remainder  of  lower  degree 
than  the  divisor  is  found  to  be  —  19  se  4-  16.  Hence  19  x  —  16  is  the  first 
Sturmian  Function.  Similarly,  the  next  remainder  is  found  to  be  —  2025. 
Hence  2025  is  the  second  Sturmian  Function. 

556.  Notation.  As  the  first  member  of  an  equation  is  repre- 
sented by  f(x)  and  its  first  derivative  by  f(x),  we  shall  represent 
Sturm's  Functions  by  fix),  f/x),  fix),  etc. 

Applying  this  notation  to  the  above  example,  we  have 

f{x)  =x*-4x2-x  +  4, 
f(x)=Sx*-$x-l, 
Mx)=  19* -16, 
/2(x)=2025. 

557.  Iff(x)  and  fix)  have  no  common  factor,  the  last  Sturmian 
Function  does  not  contain  x ;  if  they  have  a  common  factor,  the 
last  Sturmian  Function  is  their  h.  c.  d. 

558.  Sturm's  Theorem.  If  in  fix),  /'(#)?  and  Sturm's  Functions 
two  different  numbers  be  substituted,  the  difference  in  the  number 
of  variations  in  the  two  cases  ivill  be  the  number  of  real  roots  of 
fix)  =  0  that  lie  between  the  substituted  numbers,  multiple  roots  being 
counted  but  once. 

Dem.     I.    When  f(x)  =  0  has  no  equal  roots. 

1st.  Two  consecutive  functions  cannot  vanish,  i.e.,  become  0,  for 
the  same  value  of  x,  and,  consequently,  cannot  change  signs  simul- 
taneously. 

Let  the  several  quotients  in  the  process  of  obtaining  Sturm's 
Functions  be  represented  by  q,  q',  q",  q'",  etc. ;  then  by  the 
principles  of  division  we  have 

/(*)=/*(*)?-/,(*),  a) 

f(x)=Mx),,'-Mx),  (2) 

/i(*Wt(*)9"  -/.(*),  (3) 

Mx)=f3(x)q'"-Mx),  (4) 

etc.,         etc.,         etc. 


342  HIGHER  ALGEBRA 

Now  suppose  that  some  value  of  x  causes  two  consecutive 
functions,  as  /,.  (x)  and  f  (x),  to  vanish.  Then  from  (3),  f  (x)  =  0 ; 
from  (A),f4(x)  =  0;  and  so  on  to  the  last  function.  But  the  last 
function  does  not  contain  x  (Art.  557),  and  cannot  vanish  for 
any  value  of  x.  Hence  two  consecutive  functions  cannot  vanish 
for  the  same  value  of  x,  and,  consequently,  cannot  change  signs 
simultaneously.* 

2d.  The  changing  of  sign  {for  different  values  of  x)  of  any  function 
after  first,  f(x),  has  no  effect  on  the  number  of  variations. 

The  last  function  cannot  change  sign  for  any  value  of  x,  as  it 
does  not  contain  x,  and  the  other  functions  can  change  signs  only 
by  passing  through  0.  Now,  when  any  function,  as  f2(x),  becomes 
0,  (3)  becomes  f(x)  =  —  f3(x),  i.e.,  the  adjacent  functions  have 
opposite  signs,  and,  by  1st,  neither  of  these  can  change  sign  when 
f,{x)  changes.  Hence,  if  for  a  value  of  a;  a  little  less  than  that 
which  makes  f,{x)  =  0  the  signs  Of  f{x),  f2(x),  and  f3{x),  in  order, 

are  +  -\ ,  then  for  a  value  a  little  greater  the  signs  in  order 

are  H — ,  giving  only  one  variation  in  either  case. 

3d.  One  variation,  and  only  one,  is  lost  when  f(x)  vanishes  for 
increasing  values  of  x,  i.e.,  when  x  increases  through  a  root  of 

f(x)  =  0. 

Taking  values  a  little  less  and  a  little  greater  than  x  and  devel- 
oping by  Taylor's  Formula,  using  the  form  of  Art.  452,  we  have 

f(x  T  h)  =f(x)  T  f\x)  h  +f"(x)  A'  T  /"'(*)  |  +  etc. 

Let  a  be  a  root  of  f{x)  =  0.  Then  f{x)  will  vanish,  and  we 
shall  have 

f(a  Ti)=T /'(«) '*  +/"(«) f  T /'"(«) j|  +  etc. 

By  taking  h  small  enough,  the  second  member  will  have  the 
same  sign  as  its  first  term.     Hence  when  x  is  a  little  smaller  than 

*  A  function  can  change  sign  only  by  passing  through  0  or  <x>.  As  the 
functions  under  consideration  are  integral  functions  of  as,  they  cannot  become 
oo  for  any  finite  value  of  a. 


SOLUTION  FOR  INCOMMENSURABLE  ROOTS  343 

a  root  a,  /(a  —  h)  and  f'(a)  have  opposite  signs,  and  when  a;  is  a 
little  greater  than  a  root  a,  f(a  +  h)  and /'(a)  have  the  same  sign ; 
i.e.,  there  is  a  loss  of  one  variation  when  x  increases  through  a 
root  of  f(x)  =  0. 

Now,  as  there  is  a  loss  of  one  variation  whenever  x  increases 
through  a  root  of  f(x)  =  0,  causing  f(x)  to  change  sign,  and  no 
change  in  the  number  of  variations  for  the  change  of  sign  of  any 
of  the  other  functions,  the  difference  in  the  number  of  variations 
when  any  two  numbers  are  substituted  for  x  will  be  the  number 
of  real  roots  between  these  numbers. 

II.   When  f(x)  =  0  has  equal  roots. 

When  f(x)  =  0  has  equal  roots,  the  h.  c.  d.  off(x)  andf\x),  ivhich 
contains  them  (AH.  545),  is  a  factor  of  all  the  functions,  and,  conse- 
quently, does  not  affect  the  number  of  variations  obtained  from  those 
functions. 

Since  a  divisor  of  two  quantities  is  a  divisor  of  their  difference 
(Art.  110),  equation  (1)  shows  that  the  h.  c.  d.  of  f(x)  and/'(#)  is 
also  a  divisor  of  f(x)  ;  then  equation  (2)  shows  that  it  is  a  divisor 
of  f2(x)  ;  then  equation  (3)  shows  that  it  is  a  divisor  of  f(x) ;  and 
so  on.  If  the  exact  value  of  a  multiple  root  should  be  substituted, 
the  h.  c.  d.  would  be  0,  and  all  the  functions  would  vanish ;  but 
if  the  h.  c.  d.  is  +  for  any  particular  value  of  x,  its  presence  in 
all  the  functions  will  not  affect  the  signs,  and  if  it  is  — ,  it  will 
change  all  the  signs,  thus  making  no  change  in  the  number  of 
variations.  But  since  the  Sturmian  Functions  terminate  with 
the  h.  c.  d.  (Art.  557),  the  number  of  them  will  be  less  by  the 
degree  of  the  h.  c.  d.  than  it  would  be  if  there  were  no  multiple 
roots'  and  the  division  were  continued  until  a  numerical  re- 
mainder should  be  reached.  Now  the  number  of  times  a  multiple 
root  is  repeated  in  the  h.  c.  d.  is  one  less  than  in  the  original 
equation  (Art.  545).  Hence  the  multiple  roots  occur  but  once 
each  in  the  other  factor  of /(#),  and  the  difference  in  the  number 
of  variations  when  any  two  numbers  are  substituted  in  the  func- 
tions will  be  the  number  of  real  roots  between  these  limits,  each 
multiple  root  being  counted  but  once.  The  h.  c.  d.  will  give  the 
number  and  also  the  value  of  the  equal  roots  (Art.  545). 


344 


HIGHER  ALGEBRA 


559.  Cor.  Tlie  number  of  variations  lost  as  x  increases  from 
—  oo  to  0  is  the  number  of  negative  roots,  and  the  number  lost  as  x 
increases  from-  0  to  +  oo  is  the  number  of  positive  roots;  while  the 
number  lost  as  x  increases  from  —  oo  to  +  oo  is  the  total  number  of 
real  roots. 

560.  Sch.  1.  When  —  oo  or  +  oo  is  substituted,  the  sign  of  any 
function  is  the  same  as  the  resulting  sign  of  its  first  term. 

561.  Sch.  2.  As  only  the  sign  of  the  last  function  is  used 
when  there  are  no  equal  roots,  the  numerical  value  need  not  be 
found. 

EXAMPLES    CXXXVI 

Find  by  Sturm's  Method  the  number  and  situation  of  the  real 
roots  of  the  following : 


1.   a4-8ar5  +  19ar2-12a;  +  2  =  0. 

The  reason  for  applying  to  this  example  Sturm's  Method  instead  of  the 
method  in  Art.  541  is  this  :  All  the  terms  of  /(—  x)  are  -f ,  showing  that  the 
equation  has  no  negative  roots ;  and  evaluating  /(x)  for  positive 
numbers,  the  results  are  as  in  the  margin,  leaving  it  uncertain 
whether  the  roots  are  all  imaginary,  two  real  and  two  imaginary, 
or  all  real.  We  only  know  that  if  the  roots  are  all  real,  two  of 
them  lie  between  0  and  1,  and  two  of  them  between  3  and  4,  as 
these  numbers  come  the  nearest  to  satisfying  the  equation.  Evalu- 
ating for  the  tenths  between  these  numbers  would,  in  this  case, 
remove  the  uncertainty,  but  we  could  not  know  that  beforehand. 

The  application  of  Sturm's  Method  is  as  follows  : 

f(x)  =  x4  -  8  x3  +  19  x2  -  12  x  +.  2, 
i/'(x)  =  2  x3  -  12  x2  +  19  x  -  6, 
/i  (x)  =  5  x2  -  20  x  +  8, 
/2(x)=x-2, 
/s  0*0  =12. 


X 

/CO 

/'CO 

/iCO 

/«(*) 

Mx) 

Variations 

—  CO 

+ 

— 

+ 

- 

+ 

4 

0 

+ 

- 

+ 

- 

+ 

4 

+  CC 

+ 

+ 

+ 

+ 

+ 

0 

X 

/(*) 

0 

+    2 

1 

+    2 

2 

+    6 

3 

+    2 

4 

+    2 

5 

+  42 

SOLUTION  FOB  INCOMMENSURABLE  ROOTS 


345 


The  loss  of  4  variations  between  0  and  x>  shows  that  all  the  roots  are  real 
and  that  they  are  all  positive.     Again, 


X 

/GO 

/'(*) 

ZiOO 

/a  GO 

/«(*) 

Variations 

<r 

+ 

— 

+ 

- 

+ 

4 

i 

+ 

+ 

- 

- 

+ 

2 

2 

+ 

- 

- 

0 

+ 

2 

3 

+ 

0 

- 

+ 

+ 

2 

4 

+ 

+ 

+ 

+ 

+ 

0 

Hence  there  are  two  real  roots  between  0  and  1,  and  two  real  roots  be- 
tween 3  and  4. 

Evaluating  for  the  tenths  between  these  limits,  the  roots  are  found  to  lie 
between  .2  and  .3,  .5  and  .6,  3.4  and  3.5,  3.7  and  3.8.  As  the  roots  are  now 
separated,  the  remaining  figures  are  found  by  Horner's  Method. 

2.  x*  +  6x2+10x-l=0. 

3.  x4-2ar3-6ar9  +  10x  +  5=0. 

4.  z4-2ar3-8ar>+12z  +  12  =  0. 

5.  z4-2arJ  +  4x-4=0. 

6.  2xJ-15a;4  +  28ar'  +  9arJ-64x  +  42  =  0. 

7.  ^-8ar5  +  18x4-2x3-49x2  +  78x-42=:0. 

8.  a*-4ari-llx4  +  46ar}-10ar2'-76a;  +  56  =  0. 

9.  x6-  2ar>  -5a?  +  &a?  +  8a?  -8^-4  =  0. 


Solution 

/(x)  =  x6-2x5-5x4  +  8x3  +  8x2-8x-4, 

]  /'(x)  =  3  x5  -  5  x4  -  10  x3  +  12  x2  +  8  x  -  4, 

/i(x)  =  10  x4  -  13  x3  -  30  x2  +  26  x  +  20, 

/2(x)  =  27  x8  -  10  x2  -  54  x  +  20, 

/3(x)=x2-2, 

/4(x)  =  0. 

As  the  last  remainder  is  0,  /(x)  and/'(x)  have  a  common  divisor,  and  the 
given  equation  has  multiple  roots  (Art.  545).     As  the  h.  c.  d.,  x2  —  2,  occurs 


346 


HIGHER  ALGEBRA 


in  every  function,  its  presence  will  make  no  change  in  the  number  of  varia- 
tions produced  by  these  functions  (II.) .  Hence  we  evaluate  the  functions 
as  they  stand. 


X 

/(*) 

/'(*) 

/i(«) 

/*(*) 

/«(*) 

Variations 

—  CO 

+ 

- 

+ 

— 

+ 

4 

0 

— 

- 

+ 

+ 

- 

2 

+  co 

+ 

+ 

+ 

+ 

+ 

0 

-  1 

+ 

+ 

- 

+ 

- 

3 

-2 

+ 

- 

- 

- 

+ 

2 

1 

- 

+ 

+ 

- 

- 

2 

2 

- 

- 

+ 

+ 

+ 

1 

3 

+ 

+ 

+ 

+ 

+ 

0 

Hence  the  separate  roots  (each  multiple  root  being  counted  but  one)  are 
between  0  and  —  1,  —  1  and  —  2,  1  and  2,  2  and  3. 

Placing  the  h.  c.  d.  of  f{x)  and/(x)  equal  to  0  (Art.  545),  we  have 

x2  -  2  =  0, 

whence  x  =  ±  a/2. 

Hence  V2  and  —  V2  are  each  double  roots.     These  are  the  roots  that  lie 
between  1  and  2,  and  —  1  and  —  2. 

10.  a6 -4  s5  +  12arJ-3ar2-8a;-l  =  0. 

11.  x6  —  2s5-4s4  +  12ar5-3a;2-l8a;  +  l8  =  0. 

12.  a;6-  2  a5  -  2  a!4  +  12  s3  -15  a;2  -18a? +  36  =  0. 

13.  ic6-2ar5-8a;4  +  8arJ  +  20s2-8a;-16  =  0. 


RECURRING  OR   RECIPROCAL  EQUATIONS 

562.  A  Recurring  or  Reciprocal  Equation  is  an  equation  in  the 
normal  form  (Art.  510)  in  which  the  coefficients  equidistant  from 
the  two  ends  are  numerically  equal,  the  corresponding  coefficients 
having  either  all  like  or  all  unlike  signs. 

Thus,  x*  -  3  xs  +  4  x2  -  3  x  +  1  =  0, 

3x5-2x*  +  5x3-5x2  +  2x-3  =  0, 
Axn  +  B&-1  +  Cxn~2  +  .-•  +  Cx2  +  Bx  +  A  =  0, 
are  recurring  or  reciprocal  equations. 


RECURRING   EQUATIONS  347 

563.  It  is  evident  that  when  the  corresponding  coefficients  of  a 
recurring  or  reciprocal  equation  of  even  degree,  having,  therefore, 
an  odd  number  of  terms,  have  unlike  signs,  the  coefficient  of  the 
middle  term  is  0 ;  i.e.,  the  middle  term  is  wanting. 

564.  Theorem.  The  reciprocal  of  any  root  of  a  recurring  equa- 
tion is  also  a  root. 

Dem.     If  a  satisfies  the  equation 

Axn  +  Bx*-1  +  Oaf1"2  -\ \-  Cx2  +  Bx  +  A  =  0, 

-  will  also  satisfy  it :  for  the  substitution  of  the  former  gives 
a 

Aan  +  Ban-1  +  Can~2  +  ».  +  Ca2  +  Ba  +  A  =  0, 

and  the  substitution  of  the  latter  gives 

which,  when  cleared  of  fractions,  becomes 

A  +  Ba  +  Ca2  +  •••  +  Co—8  +  J^a'4"1  +  Aan  =  0, 

and  this  is  the  same  as  the  equation  obtained  by  substituting  a. 

Sch.  It  is  on  account  of  this  reciprocal  relation  of  the  roots  of 
recurring  equations  that  they  are  called  also  Reciprocal  Equations. 

565.  Theorem.     A  recurring  equation  of  odd  degree  has  -f  1  or 

—  1  as  a  root,  according  as  the  cori'esponding  coefficients  have  unlike 
or  like  signs. 

Dem.  When  the  corresponding  terms  have  unlike  signs,  the 
substitution  of  -f  1  for  x  will  cause  them  to  cancel  each  other ; 
and  when  they  have  like  signs,  the  substitution  of  —  1  for  x  will 
also  cause  them  to  cancel  each  other,  since  one  of  these  terms  is 
an  even  and  the  other  an  odd  power  of  x. 

566.  Theorem.  A  recurring  equation  of  even  degree,  ivhose  cor- 
responding coefficients  have  unlike  signs,  has  both  +  1  and  —  1  as 
roots. 

Dem.     The  equation  has  the  form 

X2"  +  Ax2"-1  4-  Bxrn  2  + Bx2  -  Ax  -  1  =  0. 


348  HIGHER  ALGEBRA 

It  is  evident  that  both  + 1  and  —  1  will  cause  corresponding 
terms  to  cancel. 

567.  A  recurring  equation  is  said  to  be  in  the  Standard  Form 
when  it  is  of  even  degree  and  its  corresponding  coefficients  have 
like  signs. 

568.  Theorem.  Every  recurring  equation  not  in  the  standard 
form  may  be  reduced  to  that  form. 

Dem.  If  the  equation  is  of  odd  degree,  it  may  be  divided  by 
X  —  1  or  x  '■+»  1,  according  as  its  corresponding  coefficients  have 
unlike  or  like  signs  (Arts.  565  and  515),  reducing  it  to  the 
standard  form. 

If  it  is  of  even  degree,  and  its  corresponding  coefficients  have 
unlike-  signs,  it  may  be  divided  by  x  + 1  and  x  —  1,  or  x2  —  1 
(Arts.  566  and  515),  reducing  it  to  the  standard  form. 

569.  Theorem.  Any  recurring  equation  in  the  standard  form 
may  be  reduced  to  an  ordinary  equation  of  half  the  degree. 

Dem.     The  equation  has  the  form 

x2"  +  Ax2"-1  +  Bx2"  2  ~.Mxn~.  Bx2  +  Ax  +  1  =  0.  (1) 

Dividing  by  xn  and  grouping  the  terms,  we  have 

(xn  +  ~V)  +  A(xn~l  +  ~hi)  +  B(xn~2  +-^\  +  —  M&  0.        (2) 

Let  x  4-  -  =  i/ ; 

x     9i 

then  (x  +1V  =  x>  +  2  +  1=  y\ 

whence      ,  x2  +  ~  =  y2  —  2. 

x2 

Similarly,  x3-\-  —  =  f  —  Sy, 

x4  +  ±  =  y4-±y2  +  2, 

X* 

xn  +  _=yn  —  nyn2-\ 

xn 


and  so  on. 
Hence 


RECURRING   EQUATIONS  349 

Now,  if  these  values  be  substituted  in  (2),  there  will  result  an 
equation  of  the  nth  degree,  which  is  half  that  of  the  given 
equation. 

EXAMPLES   CXXXVn 

Solve  the  following : 

1.   ^-11^  +  17^-f  17x2-ll^  +  l  =  0. 

Solution.  By  Art.  565,  —  1  is  a  root  of  this  equation  ;  hence  the  equa- 
tion is  divisible  by  x  +  1  (Art.  515). 

x5  -  11  x*  +  17  x:J  +  17  x2  -  11  x  +1  |_-J. 
-  12  29-12  1         0 

The  depressed  equation  is 

X4  _  12  x3  +  29  x*  -  12  x  +  1  =  0 

Reducing  to  an  equation  of  half  the  degree  by  dividing  by  x2  and  regard- 
ing x  +  -  as  the  unknown  quantity  (Art.  569),  we  have 


X2 

-12(«  +  i)=-»: 

or,  adding  2  to  both  members, 

hir- 

•  1*(*  +  I)  =  -27, 

whence  (Art.  361), 

x+-=6±3=9  or  3. 

X 

From 

x  +  -  =  9 

X 

we  have 

x  =  i(9±Vri). 

From 

x  +  -  =  3 

X 

we  have  x  =  $(3  ±  V5). 

Hence  the  roots  are  —  1,  $(9  ±  a/77),  and  J (3  ±  VE). 

2.   x4-S  ajs+4aj,-3a?+l=0.  3.   ^-5^+ 6  ^-5  x+l=0. 

4.   x4_iC3_|_a;_i  =  o.  5.   z4  +  7ar3-7a;-l=0. 

6.  x4-Ma^-19a2z2  +  4«3#  +  a4  =  0. 

7.  ax4-2x*  +  2x-a  =  0. 

a  6^-lla;4-33ar3  +  33^-f-ll«-6  =  0. 
9.  3ari-2a;4-r-5^-5x2-h2a;-3  =  0. 
10.   4x6-24z5  +  57x4-73arJ  +  57  x2-24a?  +  4  =  0. 


350  HIGHER  ALGEBRA 

570.  A  Binomial  Equation  is  an  equation  in  the  form  xn  ±  a  =  0. 
The  n  roots  of  the  equation  are  called  the  n  nth  roots  of  T  a. 

For  example,  the  solution  of  the  equation  Xs  —  1  =  0,  or  x?  =  1, 
gives  the  three  cube  roots  of  unity. 

571.  Theorem.  Every  binomial  equation  can  be  reduced  to  the 
form  yn  ±  1  =  0,  which  may  be  regarded  as  a  recurring  equation 
and  treated  accordingly. 

Dem.  In  the  form  xn  ±  a  =  0,  let  xn  =  ayn ;  then  the  equation 
becomes  ayn  ±  a  =  0,  whence  yn  ±  1  =  0. 

EXAMPLES  CXXXVIII 
Solve  the  following : 

1.   ^-1=0. 

Solution.  By  Art.  565,  1  is  a  root  of  this  equation ;  hence  the  equation 
is  divisible  by  x  —  1  (Art.  515). 

11110 
The  depressed  equation  is 

x*  +  xB  +  x2  +  X  +  1  as  0. 
Reducing  to  an  equation  of  half  the  degree  by  dividing  by  x2  and  regarding 
x  +  -  as  the  unknown  quantity  (Art.  569),  we  have 

or,  adding  2  to  both  members, 

K)2+KH- 

whence  (Art.  361), 

x  +  l  =  -±.±±V5  =  l(-l±Vl). 

x  2      2  2\  I 

Solving  this  equation  for  as,  we  have 

x  -  £(-  1  ±  V5  ±  V-  10  =F  2V5). 

These  four  roots  and  the  one  first  obtained  are  the  five  5th  roots  of  unity. 
The  5th  roots  of  any  other  number  are  these  roots  multiplied  by  the  real  5th 
root  of  that  number.  For  example,  the  five  5th  roots  of  32  (or  the  five  roots 
of  the  equation  a*5  —  32  =  0,  or  x5  =  32)  are  the  above  roots  multiplied  by  2. 


GENERAL   SOLUTION  OP  CUBlCS  351 

2.  ar*-l  =  0.  3.    ar,  +  l=0.  4.   z4-l=0. 

5.   z4  +  l=0.  6.   ar'  + 1  =  0.  7.   ^-243  =  0. 

8.   ^-1  =  0.  9.   a*  +  l=0. 

572.  From  ar3  —  1  =  (x  —  1)  (x2  +  a  + 1)  =  0,  we  obtain  for  the 
three  cube  roots  of  unity 

if  -i  +  lV-3,  -i-4V^3. 

The  two  imaginary  roots  possess  the  peculiarity  that  each  is 
the  square  of  the  other.  It  is,  therefore,  customary  to  write  as 
the  three  cube  roots  of  unity  1,  <o,  and  <o2. 

GENERAL   SOLUTION  OF   CUBICS  AND   BIQUADRATICS 

573.  Cardan's  solution  of  the  general  cubic  equation 

x3  +  px2  +  qx  +  r  =  0. 

To  transform  this  into  an  equation  lacking  the  second  power, 
we  take 

*  =  </-§  (1) 

obtaining      fy  -  f}+p(y  -  f Y+  qC V  ~  f )  +r  =  0, 

or  f  +  of  +fq  -  £\y  +  —  p3  --pq  +  r=0, 

which  has  the  form  y3  +  my  +  n  =  0.  (2) 

To  transform  this  into  an  equation  of  the  quadratic  form  (Art. 
359),  we  take 

'—J?  (3) 

obtaining  z6  +  nz3  —  —  =  0, 

which  gives  (Art.  360) 

,  n  .      hi*      m3 


'27 


II     n  ,      In2  .  m3 
=  V-  »±V4 +27- 


352  HIGHER   ALGEBRA 

Substituting  this  value  of  z  in  (3),  we  have 


*l     11        In* 


,.l2   ;    ms  t» 


27 


<V-1±>S 


r 
\4  +  27 


Multiplying  numerator  and  denominator  of  the  last  fraction  by 


n  _.    In1 


+  27' 


and  omitting  double  signs,  since  they  give  no  more  values  than  do 


single  ones,  we  have 


=  \-2  +  \4+72+^"2~\4+27' 


574.  Cor.  Cardan's  formula  fails  when  all  the  roots  are  real 
and  unequal. 

Dem.  If  a  is  one  of  the  roots,  the  other  two  may  be  found  by 
dividing  the  equation  by  x  —  a  and  solving  the  resulting  quad- 
ratic. These  roots  will,  therefore,  be  embraced  in  the  forms 
I)  _|_  Vc  and  b  —  Vc,  in  which,  for  real  roots,  Vc  may  be  either 
rational  or  surd. 

Since  the  coefficient  of  y2  is  0,  we  have  (Art.  539) 

-  a  -  (b  +  Vc)  -  (b  -  Vc)  =  0, 

whence  a  — —  2  b. 

Now  the  equation  whose  roots  are  —2  b,  b  -f  Vc,  and  b  —  Vc 

is  (kit.  520) 

2/3-(3&2  +  c)2/  +  2(&3-&c)  =  0, 

in  which  m  =  —  (3  b2  +  c), 

and  n  =  2(63-&c). 

When  these  values  are  substituted  in  Cardan's  formula,  instead 
of  obtaining  real  roots  according  to  the  hypothesis,  we  have  the 
imaginary  roots 


^-(v-bcy^ 


This  is  what  is  called  the  Irreducible  Case. 


GENERAL   SOLUTION  OF  BIQUADRATICS 


353 


575.  Sch.     Since  the  methods  heretofore  given  are  more  expe- 
ditious for  solving  numerical  cubics,  no  examples  are  appended. 

576.  Descartes1    solution   of   the    general    biquadratic    equation 
x*  -+-  ex3  +f&  +  gx  -f  h  as  0. 

Transforming  as  in  Art.  573  to  remove  the  second  term,  we 
have  an  equation  of  the  form 

yA+.nf  +  ty  +  i  =  o. 

Now  let  us  assume 


tf  +J?f  +  ty  +  I  =  (y2  +  my  +  n)  (y2  +  py  +  q)  =  0. 
Developing  and  collecting  terms, 


a) 

(2) 


V*  +jy2  +  Tcy  +  l  =  y*  +  m 
P 


f  4-  n 

y2  +  np 

mp 

mq 

Q 

+  nq. 


By  Art.  441  we  have 

m  +  p  =  0,  n  +  mp  +  q=j,  np  +  mq  =  k,  nq  =  Z, 
from  which  we  obtain 

?i=o(m2--+A  <l=7>(m2+-+j\ 

2\         m       J  2\         m       J 

Substituting  these  values  of  n  and  q  in  w</  =  I,  we  have 

m6  +  2jm*  +  (/  _  4r>m2  -  A;2  =  0. 

If  in  this  equation  we  take 

m  =  Vmj  —  }J, 

we  shall  have  a  cubic  equation,  from  which  mx  may  be  found  by 
Cardan's  formula,  and  then  m,  n,  pf  and  q  from  the  relations 
above.  Substituting  these  values  in  the  second  member  of  (2), 
and  equating  each  of  the  factors  with  0,  we  find  the  value  of  y. 


downey's  alg.  —  23 


CHAPTER   XXIV 
SERIES 

577.  Having  treated  in  Chapter  XIV  the  simpler  kinds  of 
series,  and  in  Chapter  XX  of  the  development  of  functions  into 
series  by  the  Method  of  Indeterminate  Coefficients,  Taylor's 
Formula,  and  the  Binomial  Theorem,  we  now  proceed  to  the  con- 
sideration of  series  in  general. 

SECTION  I— CONVERGENCY   OF   SERIES 

578.  We  have  seen  (Art.  438)  that  when  a  function  is  devel- 
oped into  an  infinite  series,  the  sum  of  the  series  does  not  equal 
the  function  unless  the  series  is  convergent  (Art.  434).  It  is 
clear,  therefore,  that  a  series  cannot  be  used  for  purposes  of  dem- 
onstration unless  it  is  known  to  be  convergent.  It  hence  often 
becomes  necessary  to  determine  whether  or  not  a  series  is  con- 
vergent. There  is  no  universal  test,  but  the  convergency  or 
divergency  of  series  can  usually  be  determined  by  means  of  the 
following  theorems. 

579.  Theorem.  The  convergency  or  divergency  of  a  series  is  not 
affected  by  the  addition  or  subtraction  of  a  finite  number  of  terms. 

For  the  sum  of  these  terms  is  finite  and  determinate. 

580.  Theorem.  If  a  series  all  of  whose  terms  are  positive  is  con- 
vergent, it  is  convergent  when  some  or  all  of  its  terms  are  made 
negative. 

For  the  sum  is  greatest  when  all  the  terms  are  positive. 

581.  Note.  In  what  follows,  therefore,  it  will  be  understood 
that  all  the  terms  are  positive  unless  otherwise  stated. 

582.  Theorem.  A  series  is  convergent  if  all  its  terms,  or  all  after 
a  finite  number,  are  less  than  the  corresponding  terms  of  a  series  that 

354 


CONVERGENCE  OF  SERIES  355 

is  known  to  be  convergent;  and  divergent  if  all  its  terms,  or  all  after 
a  finite  number,  are  greater  than  the  corresponding  terms  of  a  series 
that  is  known  to  be  divergent. 

The  truth  of  this  theorem  is  apparent. 

583.  To  apply  this  theorem  it  is  necessary  to  have  several 
standard  series  with  which  other  series  may  be  compared.  The 
following  serve  for  many  cases. 

I.  TJie  geometrical  series 

1+l+l+l+h+-'  (1) 

which  may  be  written  in  the  form 

i    I  *  I      *      I        *        I  *  I (2) 

2     2  •  2     2  •  2  •  2     2  •  2  •  2  •  2 

or  a??y  oMer  decreasing  geometrical  progression,  is  convergent. 
For  proof  see  Art.  329. 

II.  The  series  l-\ ^ 1 h  •••  is  convergent  when  m  >  1, 

2m     3m     4"1 

and  divergent  when  m  ^  1. 

Dem.     1st.  When  m  >  1. 

We  have  1  =  1, 

1  +  A/A 

2m     3W     2* 

etc.,  etc. 

By  adding,  we  have 

1  +  4  +  4  +  A  +  4  +  ^  +  ^;  +  etc-<1+l;  +  ^  +  etc- 

»>"»      £$"»      ^.m      5»«      (j»»      j»»  2       4 

But  this  last  series  is  a  geometrical  progression  whose  ratio  is 

2  2 

— -.     Hence,  since  m  >  1,  making  —  <  1,  the  series  is  convergent. 

2m  2m 

2d.     When  m  =  1,  giving  the  series 

^2^3^4^5^  W 


356  HIGHER  ALGEBRA 

Grouping  the  terms  thus,  • 

it  is  seen  that  each  group  is  greater  than  \.     Hence  the  series  is 
greater  than 

i +1+1+1+1+ -, 

and  is,  therefore,  divergent. 
3d.     When  m  <  1. 

In  this  case  each  term  after  the  first  is  greater  than  the  corre- 
sponding term  of  the  series 

which,  as  shown  above,  is  divergent. 


EXAMPLES  CXXXIX 

Determine   whether   the   following    series   are   convergent   or 
divergent : 

Scg.     Compare  with  (2). 

2.  1  +  3  +  -  +  -  +  -  +  -  +  -+--. 

Solution.     After  the  5th  term  each  term  of  this  series  is  less  than  the 
corresponding  term  of  the  geometrical  progression 

4-4      4-4.4      4.4.4.4 
whose  ratio  is  |.     Therefore,  by  I,  the  given  series  is  convergent. 

3.  1+1  +  1  +  1  +  1  +  .... 

T  23      33     43     53 

4^9^16  n2 

Sug.     Compare  with  (3) . 


CONVERGENCT  OF  SERIES  357 

584.  Theorem.  A  series  is  convergent  if,  from  the  beginning  or 
after  a  finite  number  of  terms,  the  ratio  of  each  term  to  the  preceding 
term  is  less  than  some  quantity  which  is  itself  less  than  1. 

Dem. 


the  series  be 

S= \-k  +  l  +  m  +  n-\ , 

a) 

=-+*H+^+-> 

(2) 

=...+fcfi+i+^+^+.. 

\        k      kl       Mm 

..). 

(3) 

Now  if  the  ratio  of  each  term  of  (1)  to  the  preceding  term  is 
less  than  p,  we  have  from  (3) 

S  <...  +  & (1+j,  +p2  +  p3  +...)• 

Hence  if  p  <  1,  we  have  from  Art.  329 

1-p 

and  the  series  is  convergent. 

585.  Cor.  A  series  is  convergent  if  from  the  beginning  or  after 
a  finite  number  of  terms,  the  ratio  of  each  term  to  the  preceding  term 
is  less  than  1  and  approaches  0  as  a  limit. 

For  the  ratio  is  then  always  less  than  a  quantity  which  is  itself 
less  than  1. 

586.  Sch.  The  series  (3)  of  Art.  583  shows  that  it  is  not 
sufficient  in  the  theorem  of  Art.  584  to  say  that  the  ratio  of 
each  term  to  the  preceding  term  is  less  than  1.  It  is  less  than 
1  in  this  series,  but  approaches  1  as  a  limit  as  the  terms  are 
indefinitely  increased. 

587.  Theorem.  A  series  is  divergent  if,  from  the  beginning  or 
after  a  finite  number  of  terms,  the  ratio  of  each  term  to  the  preceding 
term  is  equal  to  or  greater  than  1. 

For  the  sum  of  the  series  is  the  sum  of  an  infinite  number  of 
finite  terms. 


358  HIGHER  ALGEBRA 

588.    Theorem.     A  series  of  numerically  decreasing  terms  which 
are  alternately  -\-  and  —  is  convergent. 

Dem.     Let  the  series  be 

S  =  a  -  b  +  c  -  d  -f  e  -/H .  (1) 

This  may  be  written  in  the  forms 

S  =  (a-b)  +  (c-d)  +  (e  -f)  +  .-.,  (2) 

and  8  =  a  -  (b  -  c)  -  (d  -  e)  -  (/-  g) .  (3) 

From  (2)  8  >  a  -  b, 

and  from  (3)  S  <  a. 

Hence  the  series  is  convergent. 

EXAMPLES   CXL 

Determine   whether  the    following    series    are   convergent   or 
divergent. 

n&  n&  /y" 

1.   x  +  QL+°L  +  aL+  .... 
•    2T3      ± 


,  1    1,1    1,1    1  , 

2-1-2  +  3-4  +  5-6+-- 

3.   The  logarithmic  series 

x2  .  Xs     x4  .  x5 
"-2+3"4+5- 

x« 
6 

1.1.1.1. 

1.2     3-  45-  6     7-  8 

2         22         23         24  2n 

5'   rT24"2T3  +  3T4  +  475+'"n(n  +  l)"f '"' 

Solution.     The  limit  of  the  ratio  of  the  (w  +  l)th  term  to  the  nth  term  is 

2«+i  2M  2  w 

(«+l)(«  +  2)'r«(»  +  l) 

Hence  (Art.  587)  the  series  is  divergent. 

6.   1+1  +  *  +  i  +...  +  »  +  .... 
2     2*     2*     2*  2" 


-*"-1    =2 
«  +  2_L 


CONVERGENCY  OF  SERIES  359 

7.   — I 1 1 r  •  •  • 

x     1  +  x     2  +  x     3  +  x 

Sug.     Compare  with  (3),  Art.  583. 

1  11  1 

■   l  +  x     1  +  2X2     1+3^  "'"""l  +  naf 

1  _1 11 

'  l  +  a>     1  +  2*     1  +  Sa?     l  +  4a? 

10.  l+l+I+I+...-lj+---. 

2      32     43  nn  ■ 

11.  1 *__  +  — ^ ^__l...... 

1  +  a     1  +  2  ct      l+3a 

12.  The  Exponential  Series  (Napierian) 

Solution.     The  limit  of  the  ratio  of  the  (n  +  l)th  term  to  the  nth  term  is 
xn+\       xn  _x~\        _q 
jft  +  1      [n      wJ„=x 
Hence  (Art.  585)  the  series  is  convergent  for  all  values  of  x. 

13.  The  Binomial  Series 

(1  +  X)m  =  1  +  mx  +  m(m~V>x2  +  -. 
If 

m(m-l)..-(m-yi  +  2)a;W_1  |       ^ 

|m  —  1 

Solution.     The  limit  of  the  ratio  of  the  (n  +  l)th  term  to  the  nth  term  is 
m(m  —  1)  •••  (w  —  n  -f  l)xw      w(w  —  !)•••  (m  —  n  +  2)xn~1 


n  \       11  J     Jn=oo 


If,  then,  x  be  numerically  less  than  1,  the  series  will  be  convergent 
(Art.  581). 

It  may  be  shown  that  the  series  is  convergent  when  x  =  1 ,  provided 
m  >  —  1  ;  also  when  x  =  —  1,  provided  m  is  positive.  See  C.  Smith's  Trea- 
tise on  Algebra,  Art.  338. 


360  HIGHER  ALGEBRA 


i4-  1+\+i+i+- 

■l  e  ^  [         ^  j         g  I  j ^ i 

'   l.aT2\3T8v4-         *(»  +  !)* 

16.  The  exponential  series 

a"  =  1  +  (log,  a)x  +  (log,  a)2|  +  (log,  a)3j|  +  •  •  -. 

17.  The  series  whose  7*th  or  general  term  is  —  • 

\n 

18.  The  series  whose  nth  term  is 


(n  +  l)n+1 

SECTION  II  — SCALE   OF   RELATION  OF   A   RECURRING 

SERIES 

589.  A  Recurring  Series  is  a  series  in  which,  either,  from  the 
beginning  or  after  a  finite  number  of  terms,  each  term  is  equal  to 
the  algebraic  sum  of  the  products  of  a  fixed  number  of  the  pre- 
ceding terms,  multiplied,  respectively,  by  certain  quantities  which 
remain  the  same  throughout  the  series. 

590.  A  recurring  series  is  of  the  First,  Second,  Third,  etc.,  Order, 
according  as  each  term  is  derived  from  one,  two,  three,  etc.,  of  the 
preceding  terms. 

Thus,  by  the  method  of  indeterminate  coefficients  we  found  (page  275) 

1  _  3  x  -  x2  =  x  _     _  2  x2  _  g    3  _  12  xi  _  etc 

1  -2x-x2 

In  this  series  each  term  after  the  third  is  equal  to  2  x  times  the  preceding 
term  plus  x2  times  the  second  preceding  term.  Hence  it  is  a  recurring  series 
of  the  second  order. 

591.  If  a  recurring  series  of  any  order  be  written  in  the  form 
ux  +  u2  +  u3  H h  u^2  +  un_x  +  vn  +  un+1  +  un+2  +  —, 

we  shall  have  by  definition 

un  =pxun_1  +  qtfun_2  +  rx*un_s  +  ••-, 
whence  un  —  pxUn^  —  gAn_2  —  7*An_3  —  •  ••  =  0, 

which  expresses  the  law  of  the  series. 


SCALE  OF  RELATION 


361 


In  this  form 


1  —  px  —  qtf  —  rx*—-- 


constitutes  the  Scale  of  Relation,  or  the  Scale,  and  p,  q,  r,  etc.,  are 
the  Constants  of  the  Scale. 

Thus,  in  the  series  of  Art.  590  the  scale  of  relation  is  1  —  2  x  —  x2,  and  the 
constants  of  the  scale  are  2,  1. 

592.  To  extend  a  series  some  of  whose  terms  are  given,  it  is 
sufficient  to  make  use  of  the  constants  of  the  scale,  since  the 
powers  of  the  variable  may  be  supplied  by  inspection. 

593.  Prob.     To  find  the  constants  of  the  scale. 

Solution.     1st.  In  a  series  of  the  first  order. 
Let  the  series  be 

a  +  bx  +  ex2  +  dx*  +  ex4+fx6+  ••-, 
and  p  the  constant  of  the  scale. 


Then 


f=l™, 


whence 


p  = 


f 


The  series  is  a  geometrical  progression. 
2d.    In  a  series  of  the  second  order. 
Let  p,  q  be  the  constants  of  the  scale.     Then 
f=*pe  +  qd, 
e=pd  +  qc, 
from  which  p  and  q  may  be  found. 

3d.    In  a  series  of  the  third  order. 

Letp,  q,  r  be  the  constants  of  the  scale.     Then 

f=pe  +  qd  +  rc, 

e=pd  -\-qc+  rb, 

d  =  pc-\-qb  +  ra, 

from  which  p,  q,  and  r  may  be  found. 

We  may  proceed  in  the  same  way  for  series  of  higher  order. 

When  the  order  is  not  known,  we  may  assume  it  of  the  second, 
third,  etc.,  order  until  the  right  order  be  found.  If  the  order  be 
assumed  too  high,  one  or  more  of  the  constants  will  be  0 ;  and  if 


362  HIGHER  ALGEBRA 

assumed  too  low,  the  error  will  become  apparent  in  applying  the 
scale  found. 

594.   Cor.     To  find  the  scale  ive  must  have  twice  as  many  terms  of 
the  series  as  there  are  constants  in  the  scale. 

EXAMPLES    CXLI 

Find  the  constants  of  the  scale  in  each  of  the  following,  and 
extend  each  series  one  term  : 

1.   i  +  4  x  +  6  x2  +  11  x*  +  28  x*  +  63  as8  +  131  x«  +  •  •  •. 

Solution.     Assuming  the  series  of  the  second  order,  we  have 

6p  +  4g  =  ll, 

whence  P  —  H  and  Q  =  f • 

If  the  proper  constants  have  been  found,  we  shall  have 
lljp+  6g  =  28; 
but  ll(tf)+6(*)=19.7. 

Hence  the  series  is  not  of  the  second  order. 
Next  assuming  it  of  the  third  order,  we  have 

6p  +  4g  +  r  =  11, 

llj)  +  6g  +  4r  =  28, 

28p  +  Uq  +  6r  =  63, 

whence  p  =  2,  q  =  —  1,  r  =  3. 

If  the  proper  constants  have  been  found,  we  shall  have 

63j>  +  28g  +  llr  =  131. 

Now  63-2  +  28(-l)+ 11-3  =  131. 

Hence  the  proper  constants  have  been  found. 
To  find  the  coefficient  of  the  next  term  we  have 

131-  2  +  63(-l)+28-  3  =  283. 

Hence  the  next  term  is  283  x"1. 

2.  1 +  6^  +  12^  +  48a3 +  120«4+"-. 

3.  l  +  3a  +  7a2  +  17ar5  +  41x4+—. 

4.  l-f-9x-15ar94-57o3-159a4+-.-. 

5.  l  +  #  +  2a2  +  2x3  +  3a;4  +  3ar5  +  4a;6  +  .... 


THE  NTH  TERM  OF  A    SERIES  363 

6.  2  +  z-3z2  +  2ar5  +  :K4-3ar5+.... 

7.  3  +  5 a-  +  7  x>  +  13  ar3  +  23a;4  +  45*"  +  87.T6  +  •••• 

SECTION   III  — THE   NTH   TERM   OF   A  SERIES 

595.  We  have  learned  how  to  find  the  nth  term  of  an  arith- 
metical series  (Art.  318),  a  geometrical  series  (Art.  326),  the 
binomial  series  (Art.  457),  and  of  a  recurring  series  by  extension 
to  the  nth  term  by  means  of  the  constants  of  the  scale  of  relation 
(Art.  593).  A  useful  method  of  finding  the  nth  term  of  a  less 
simple  series  is  by  the  Successive  Orders  of  Differences. 

596.  The  First  Order  of  Differences  of  a  series  is  the  series 
obtained  by  subtracting  the  1st  term  of  the  given  series  from 
the  2d,  the  2d  from  the  3d,  the  3d  from  the  4th,  and  so  on.  The 
Second  Order  of  Differences  is  the  series  obtained  from  the  first 
order  as  the  first  order  is  obtained  from  the  given  series.  The 
Third,  Fourth,  etc.,  Orders  are  obtained  similarly. 

Thus,                          given  series,  1,  8,  27,  64,  125,  etc., 

1st  order  of  differences,  7,  19,  37,  61,  etc., 

2d  order  of  differences,  12,  18,  24,  etc., 

3d  order  of  differences,  6,    6,  etc., 

4th  order  of  differences,  0,  etc. 

597.  Prob.     To  find  the  first  term  of  any  order  of  differences. 
Solution.     Let  the  series  be  a,  b,  c,  d,  e,  ••• 

1st  order  of  diff.,  b  —  a,  c  —  b,  d  —  c,  e  —  d,  ••• 

2d    order  of  diff.,      c  —  2 6  +  a,      d  —  2c +  b,      e  —  2d  +  c,  ••  • 
3d    order  of  diff.,  d  -  3  c  +  3  b  -  a,  e  -  3  d  +  3  c  -  b,  ••• 

4th  order  of  diff.,  e  —  4 d  +  6c  —  46  +  a,  ••• 

Denoting  the  first  terms  of  the  respective  orders  of  differences 
by  Dlf  D2,  Z>3,  Z>4,  etc.,  we  have 

Dx  =  —  a  +  b, 

D2=a  —  2b  +  c, 

D3=-a  +  3&-3c  +  d, 

B4  =  a  -  4  b  +  6  c  —  4  d  +  e, 
etc.,  etc. 


364  HIGHER  ALGEBRA 

The  coefficients  in  these  terms  are  seen  to  be,  numerically,  those 

of  a  developed  binomial  by  the  binomial  theorem.     Hence,  when 

n  is  even, 

^                 ,    .  n  (n  —  1)        ft  (ft  —  1)  (ft  —  2)  7  , 
Dn  =  a-nb+-±— — }-c ^ -£* ld  +  •  •, 

and  when  n  is  odd, 

r»  ,     7      w  (n  —  1)     ,  ft  (ft  —  1)  (ft  —  2)  7 

Dn  =  -a  +  7ib--^— — ^c+-^ -^ *d . 

598.  Cor.     To  find  the  1st  term  of  the  nth  order  of  differences, 
n  +  1  terms  of  the  series  must  be  given. 

This  is  seen  by  inspecting  the  values  of  Dlf  D2,  D3,  D4,  etc. 

EXAMPLES  CXLII 

Find  the  first  term  of  the  specified  order  of  differences  in  each 
of  the  following: 

1.  3d  and  4th  of  7,  12,  21,  36,  62,  etc. 
Solution.     For  the  third  order  we  have 

Z>3=-«  +  3&-3c  +  d  =  -7+3.12-3.21+36  =  2. 
For  the  4th  order, 

Z)4  =  a-46  +  6c-4d+e  =  7-4.12  +  6.21-4.36  +  02  =  3. 
In  practice  the  shortest  way  is  to  find  the  successive  orders  by  subtraction. 

2.  3d  of  1,  3,  6,  10, 15,  etc. 

3.  3d  and  4th  of  1,  8,  27,  64,  125,  etc. 

4.  3d  and  5th  of  1,  3,  32,  33,  34,  35,  etc. 

599.  Prob.     To  find,  by  the  successive  orders  of  differences,  the  nth 
term  of  a  series. 

Solution.     From  Art.  597  we  obtain 
b  =  a  +  Dl} 
c  =  a  +  2Dl  +  D2, 
d  =  a  +  3Dl  +  3D2  +  D3, 

e  =  a  +  4A  +  6A  +  4A  +  A, 
etc.,  etc. 


SUMMATION  OF  A    SERIES  365 

It  is  seen  that  the  coefficients  of  the  nth  terra  of  the  series  are 
the  coefficients  of  the  (n  —  l)th  power  of  a  binomial.  Hence, 
writing  n  —  1  for  n  in  the  coefficients  of  the  binomial  formula 
(Art.  457),  we  have 

nth  term  =  a  +  (* U 1)  A  +  (w~1Hn~2)Z>> 

j  (rc-l)(n-2)(m-3)A| 

It  is  evident  that  the  nth  term  of  a  series  can  be  found  exactly 
only  when  the  terms  of  some  order  of  differences  are  0. 

EXAMPLES  CXLIH 

Find  the  terms  specified  in  the  following : 

1.  12th  term  of  1,  5,  15,  35,  70,  126,  etc. 

Solution.     Obtaining  the  successive  orders  of  differences,  we  have 

4,  10,  20,  35,  56,  etc.,  whence  DL  =  4 

6,   10,  15,  21,   etc.,     whence  D2-Q 

4,    5,     6,  etc. ,         whence  Dz  =  4 

1,     1,  etc.,  whence  Z>4  =  1 

0,  etc.,  whence  Z>5  =  0. 

Substituting  in  the  formula,  we  have 

„tUorm      !    .  „     ,   .  1L10A  .  11-10-9.   ,  11 -10.9-8      1Q(., 

nth  term  =  1  -f  11  .  4  -\ 6  -\ 4  H =  1365. 

2  2-3  2-3-4 

2.  12th  term  of  1,  3,  6,  10,  15,  21,  etc. 

3.  15th  term  of  1,  22,  32,  42,  etc. 

4.  12th  terra  of  1,  4z,  6a,-2,  liar5,  28a;4,  633^,  etc. 

SECTION  IV  — SUMMATION  OF   SERIES 

600.  We  have  learned  how  to  find  the  sum  of  n  terms  of  an 
arithmetical  series  (Art.  319),  and  of  a  geometrical  series  (Art. 
327),  and  the  limit  of  the  sura  of  an  infinite  decreasing  geometrical 
series  (Art.  329).  We  proceed  to  develop  methods  for  finding 
the  sum  of  series  that  are  less  simple,  though  there  is  no  general 
formula  for  summation. 


366  HIGHER   ALGEBRA 

601.  The  sign  of  summation  is  the  Greek  letter  sigma,  X 
Written  before  the  general  term  of  a  series  it  signifies  the  sum  of 
the  series  obtained  by  making  n  successively  equal  to  1,  2,  3,  4,  etc. 

Thus,  V 1 =  _!_  +  _!_  +  _J_ +  _!_+.... 

4n(n  +  l)      1-2      2-3     3- 44- 5 

602.  The  Generating  Function  of  a  series  is  the  function  which, 
when  expanded  (Art.  436),  produces  the  series. 

Thus,  since   *  ~  3  x  ~  x*  =  1  -  X  -  2  x2  -  5  x3  -  12  x1 


1  -  2  x  -  x2  1  -  2  x  -  x2 

is  the  generating  function  of  the  series  1  —  x  —  2  x2  —  5  x3  —  12  x4  —  •••. 

The  generating  function  of  an  infinite  series  is  the  same  as  the 
sum  of  the  series  when  the  series  is  convergent,  but  not  when  the 
series  is  divergent  (Arts.  436  and  438). 

603.  Prob.  To  find,  by  the  method  of  decomposition,  the  sum  of 
a  series  whose  general  term  has  the  form 


q  q 

-—± or y 

n(n  +p)         n(n  +  p)  (n  +  2  p) 


or,  in  general, 


n  {n  +  p)  («  +  2  p)  •  •  •  (n  +  rp) 
Solution.     By  Art.  447,  we  have 

q         =A+     B 


n  (n  -f  p)      n      n  +  p 

1  1 

whence  A  =  -  and  B  = 

p  p 


Hence 


p\n      n+  p 


n(n+p) 

,  y q_  =yv?__i_vvy?_y_L_>i  m 

L*n{n+p)      L*p\n      n+p)     p\L*n     L^n+pJ 

Now,  when  n  is  taken  successively  equal  to  1,  2,  3,  4,  etc.,  all 

the  terms  of   /   -  and   /   -^—,  except  the  first  p  terms  of  the 
L*n  L*n+p 


SUMMATION  OP  A   SERIES  367 

former  and  the  last  p  terms  of  the  latter,  will  cancel,  and  the  sum 

will  be  -th  the  sum  of  these  remaining  terms.     Hence 
P 

y   g 

L^nin  +  p) 

=  -1 1st  p  terms  of   /  -  —  last  p  terms  of   /  — - —  ).     (2) 
p\  —>  Amjn  +  pJ 

The  sum  of  an  infinite  number  of  terms  is  -th  the  sum  of  the 

Zq                    Q       1  P 

-,  since  — y-2 r      =  0. 
n             *{n+P)X 

Similarly,  we  have 

Zg        _  i  rsp    g      y       q       v  (3) 
n(n+i>)(u  +  2p)     2p\L*n(n+p)     A*(n+p)(n+2p))' 

and,  in  general, 

? =-(Y <? 

n  (n  +p)  (?i  +  2  p)  •  •  •  (n  +  rp)      rp  \L**  n  (n  +p)  •  •  •  [ n  +  (r — l)p] 

_y 2 v  (4) 

EXAMPLES  CXLIV 

Find,  by  the  method  of  decomposition,  the  sum  of  the  follow- 
ing to  n  terms  and  to  oo. 

1.22.3      3.44.5  n(n  +  l) 

Solution.     Since  q  —  1  and  p  =  1,  we  have  from  (2) 

5] 2 1st  term  of  ^  -  -  nth  term  of  ^J  — — 

**  n  (»  +  p)  ^  n  T»  n  +  1 

Hence  A,  - 


1      n  +  1      w  +  1 

and  &o  as  1st  term  of  5/-  — 1« 

The  value  of  Sx  is  also  obtained  from  the  value  of  Sn  by  making  n  =  00, 
giving 


n  +  Uoo 


368  HIGHER  ALGEBRA 

Without  the  use  of  the  formula  we  may  proceed  thus  i 

%    _1     1       1    _\      1       1    _1     1 

1-2      1      2     2  •  3      2      3'    3  •  4      3      4     G  °' 

Hence  the  series  becomes 

(i_i\  +  /l_l\  +  /i_i\+...+/l__l_\  =  ,__J_  =  _S_. 

i  .32.43.5 

Solution.     Since  q  =  1  and  p  =  2,  we  have  from  (2) 

5) 2 1/  ist  2  terms  of  V  -  -  last  2  terms  of  T  -J—  V 

^w(»+p)     2V  ^n  ^n  +  2j 

Hence  «,  =  1(1  +  1^I__!_U1(8_1  __1_\ 

2\1      2      n     n  +  «y     2\2     n     n  +  2/ 

and  ^  =  i(?-X— Ul    =f. 

"     2\,2      n     «,+  2/J„     4 

3.       1     +     1     +  _1_  +     1     +...+.        1 


1-42.53-64.7  «  (n  +  3) 

3-8      6,12     9-16 


Sug.     Write  in  the  form 


i2Vi-2   2.3   3.4      y 

and  compare  with  ex.  1. 


5  1      1      1      ■      1      I       1       . 

'   1-4     2-6     3-8     4-10  ' 

6  1      I      1       I      1      I    -     I  1  I     • 

'   1-3      3.5^5.7  (2n-l)(2n  +  l)  ' 

Solution.     As  (2)  is  not  applicable,  the   denominator  not  having  the 
form  n(n  +  J)),  we  have  from  (1), 

V 2 =1(2—1 T  — L_ ) 

^(2«-l)(2«+l)      2V-^2«-l      -W2n+1/ 

=1/1+1+1+. ? 1_I ! L_\. 

2V1     a     6         2)1-1      3     5  2n-l     2n  +  1/ 


SUMMATION  OF  A    SERIES  369 

Hence  Sn  =  \{  1  -  ^— A  =  — -^— -» 

2\        2n  +  l/      2n  +  1 

and  ^=2^nL=i- 

.      2         2  2  2 

'   3.5"h5.7"t"7.9        *~1~(2tt  +  l)(2n  +  3)i~'"' 

8.    i+2      3  _^_       ^ 

1         1 


\n  +  1      [n     l/i  +  l 

9_2 3,  _4 SL.+  ...:± ""  +  !  t 

3-5     5-7      7-9     9-11  (2n  +  l)(2n  +  3) 

Solution.    Since  5  =  n  +  1,  a  variable,  andp  =  2,  we  have  from  (1), 

A(2n+l)(2n  +  3)      2\^f2n  +  l      -W2n  +  3/ 

=  y2_3  ,  4_#  n  +  1       2      3  71        ,    ^  +  1  \ 

2\3      5      7  2n+l      5     7  2w  +  l      2n  +  3/ 

2V3  2n  +  3J 

Hence,  when  w  is  even, 

q  -V2      11    "+1  "UV      1  1    w  +  1  V 
n"2\3  2W  +  3J      2\      3"^2«  +  3/' 

and  when  n  is  odd, 

*  =l/2       n  +  l\ 
n     2\3     2w+3y/ 

1  1  1  1  . 

'   1.2.3"t"2.3.4"t"3.4.5_h'  '      n(w  +  l)(w  +  2)  ^  ' 
Sue.    Use  (3). 

1.3.5^3.5.7^5.7.9^ 


12.     T"^^  +  ^-4-T  + 


1-2.3        2.3.4        3.4.5 

downey's  alg.  — 24 


370 


HIGHER  ALGEBRA 


Solution.    Since  q  =  n  +  3  and  p  =  1,  we  have  from  (3) 

y       g       =iiy w+8   y    n+3    ^ 

^f*(n+p)(n  +  2p)     2\^n{n  +  \)     <W(W  +  i)(n  +  2)/ 
1/    4_  ,  _5_      _6_  .      7i  +  3 

2     2-3      3-4      ""      »(n+l) 


-V- 


4 5_  >i+2 n  +  3         \ 

2-3     3-4  »(»+I)      (n  +  l)(n  +  2)j 

=  I/r2+-J-  +  J-+...+— i w  +  3         ^ 

2\        2-3     3-4  w(n  +  l)      (n  +  l)(n  +  2)/' 


and 
13. 

14. 


.c 

■** 

1 
2" 

1 

n  +  1 

w  +  3 

(n  +  l)(n 

+  2) 

=  !(«- 

2V2 

c« 

2n  +  5        \ 

#00 

_5 
4' 

+ 


+ 


+ 


1.2.32.3.43.4.54.5-6 

3.4.54.5.65.6.7 


+ 


15. 


+ 


+ 


+  .... 


1-3     1.3.5     1.3.5.7  1.3.5...(2n  +  l) 


+  •••• 


604.    Prob.     To  find,  by  the  scale  of  relation,  the  sum  of  n  terms 
of  a  recurring  series. 

Solution.     Let  the  series  be  a +- bx -\-  ex2  +■  dar3  +-  •••, 

whose  scale  of  relation  (Art.  591)  is  1  —  px  —  qx2. 

Though  this  assumes  that  the  series  is  of  the  second  order,  the 
method  is  general. 

Representing  by  Sn  the  sum  of  n  terms,  we  have 

Sn  =  a  +-  bx  +-  ex2  +-  dx*  -\ h  l®"'1- 

Multiplying  both  members  by  the  scale  of  relation  and  arrang- 
ing according  to  the  powers  of  x,  we  have 


(1—  px— qx2)Sn=a+  b 
—pa 


X+-    c 

x'2+-  d 

—pb 

—pc 

—  qa 

-qb 

x?-\ h     l\xn~l 

—pk\       —pi 

-  qj\     —g* 


-qlxn+\ 


SUMMATION   OF  A   SERIES  371 

Since,  by  Art.  591,  the  coefficients  of  a2,  ar3,  •••#n~1  are  0,  we 
have 


(1  —  px  —  qx2)  Sn  =  a+    b 
—  pa 


x—  pl\  xn 
—  qk\      —  qlxn+l. 


Hence     Sn  =  a  +  (b  ~ pa)  x  ~  ^  +  ^)af  ~  qlx"+1 

1  —  px  —  qx2 

605.  Cor.     The  sum  of  an  infinite  convergent  recurring  series  of 
the  second  order  is 

&   _a+(b  —  pa)x 
1  —  px  —  qxr 

This  is  because  the  last  two  terms  of  the  expression  for  Sn 
approach  0  as  a  limit. 

606.  Sch.     The  expression 

a  +  (b  —  pd)x 
1—  px—  qx2 

is  the  generating  function  (Art.  602),  and  the  development  of  it 
will  reproduce  the  original  series. 

EXAMPLES  CXLV 

Find,  by  the  scale  of  relation,  the  generating  function  of  each  of 
the  following : 

1.   l  +  2»  +  8a?l  +  28aj8  +  100aJ4  +  .... 

Solution.     We  must  first  find  the  constants  of  the  scale.     From  Art.  593 
we  have 

S  =  2p  +  q, 

2S  =  8p  +  2q, 

whence  p  =  3,  q  =  2. 

Since  28  p  +  Sq  =  100, 

the  proper  constants  have  been  found. 

Substituting  in  S  =  « +(&-!»)* 

I  —  px  —  qx2 

we  have  S= 

1  -  3  x  -  2  x2 


372  HIGHER  ALGEBRA 

2.  l  +  9a-15ar  +  57ar3-lo9a;4+  .... 

3.  l  +  2«  +  3ar*  +  5ar3  +  100£4+---. 

4.  1 +5a  +  9ar9  +  13arJ+ .... 

5.  1 +  3x  + 8^  +  22 ar^  60a4 +•••• 

6.  2  -  5a  +  17a2 -65^  + 257a4 . 

7.  1  +  3&*-{-o£2  +  7£3H . 

8.  3  +  5a  +  7x2  +  13a3  +  23a4  +  45a5  +  •••. 

9.  1  +  3  a  —  x2  —  5a3  —  7  a4  —  a^  +  lla^H . 

10.   1  -  3  x  +  5  x2  +  5  a8  +  13  x4  +  61  a5  +  181  a6  +  •  •  •. 

607.  Prob.  To  Jfndj  &#  £7ie  method  of  differences,  the  sum  of  n 
terms  of  a  series. 

Solution.     Let  the  given  series  be 

a,  b,  c,  d,  e,  f  etc.,  (1) 

and  let  S  represent  the  sum  of  n  terms. 

Then  8  is  the  (n  +  l)th  term  of  the  series 

0,  a,  a  +  b,  a  +  b  -+-  c,  a  +  b  +  c  +  d,  etc.  (2) 

Series  (1)  is  the  same  as  the  first  order  of  differences  (Art.  596) 
of  series  (2),  the  first  order  of  differences  of  (1)  is  the  same  as  the 
second  order  of  differences  of  (2),  and  so  on. 

Substituting  in  the  formula  for  the  nth.  term  (Art.  599)  n  + 1 
for  n,  0  for  a,  a  for  D1}  Dx  for  D2,  etc.,  we  have 

c             .  n  (n  —  1)  7-v    .  n  (n  —  1)  (n  —  2)  ^    .     . 
S  =  na-\ — ^— — L  Dx  H — * -^ — i — L  D2  +  etc. 

This  formula  is  applicable  only  when  the  series  is  such  that  all 
the  terms  of  some  order  of  differences  become  0. 

EXAMPLES  CXLVI 

Find,  by  the  method  of  differences,  the  sum  of  the  following : 
1.   1,  8,  21,  40,  65,  etc.,  to  12  terms. 

Solution.     Obtaining  the  successive  orders  of  differences,  we  have 

1,  8,  21,  40,  65,  ... 

7,  13,  19,  25,  - 

6,    6,    6,  -. 

0,  0,  ... 


PILES  OF  SPHERICAL   SHOT  373 

Substituting  in  the  formula, 

5=12xl  +  1^12-1^  +  12(12-,1K12-2)6  =  1794. 

2.  1,  3,  5,  7,  etc.,  to  20  terms. 

3.  4,  14,  30,  52,  80,  etc.,  to  13  terms. 

4.  1,  2,  3,  4,  5,  etc.,  to  50  terms. 

5.  1,  5,  15,  35,  70,  126,  etc.,  to  30  terms. 

6.  7,  14,  19,  22,  23,  etc.,  to  9  terms. 

7.  The  following,  obtained  as  above,  are  useful  in  Physics : 


2n  =l+2  +  3  +  4H \-n 


(n  +  1) 


2       ' 

2n2  =  l2  +  22  +  32  +  42  +  •••  +  n2  =  n(n  +  1^2n  +  1\ 

6 

2n3  =  l3  +  23  +  33  +  43  +  ...  +n3  =  n2(n  +  *)*  =  (2n)2, 

Sn4  =  14  +  24  +  34  +  44+...  +  n4  =  ^  +  l)(67i3  +  9n2  +  n-l) 

If  n  is  a  large  number  and  only  approximate  results  are  sought, 
all  the  lower  powers  may  be  omitted,  and  we  have 

*-!*   *•-$  Sn«~f,  **•-£ 

2  3  4  5 

and,  in  general, 

^  »       wm+1 


ra  +  1 


^n(n  +  l)      1-2      2.3     3-44.5  n(n+l) 

2  2"r2~h2~t~2~r  2 

=  i[2  (n2  +  n)]  =  \  (Sn2  +  2M)  =  »  ("  +  ff"  +  2)- 

PILES   OF   SPHERICAL   SHOT 

608.  Prob.  To  find  the  number  of  balls  in  a  complete  pyramid 
or  wedge. 

Solution.     1st.   A  triangular  pyramid. 

Let  n  be  the  number  of  balls  in  one  side  of  the  bottom  course. 
This  will  also  be  the  number  of  courses. 


374  HIGHER  ALGEBRA 

The  number  of  balls  in  the  successive  courses,  beginning  at  the 
top,  is 

1,  3,   6,   10,   15,   etc. 

Obtaining  the  successive  orders  of  differences  and  substituting 
in  the  formula  of  Art.  607,  we  have 

o  _n(n+l)(n  +  2) 

2d.   A  square  pyramid, 

The  number  of  balls  in  the  successive  courses  is 

l2,   22,   32,   42,    52,  ...  n\ 
As  before,  we  obtain 

o      w(n  +  l)(2n  +  l) 

B 

#d   ^4  wedge  ivith  rectangular  base  ayid  single  row  at  top. 

Let  m'  be  the  number  in  the  top  row.  Then  the  next  course, 
being  longer  by  1,  will  contain  2(m'  +  l);  the  next,  3(m'+2); 
the  next,  4  (m'  -f  3),  and  so  on,  giving  the  series 

m',  2  m' +  2,  3m' +  6,  4m' +  12,  .... 
As  before,  we  obtain 

s  _  n(n  +  1)  (3  m'  +  2  n  -  2) 
6 
If  we  let  m  be  the  number  of  balls  in  the  length  of  the  base, 
we  have 

m'  =  m  —  n  -f  1, 

and  the  last  formula  becomes 

p  _  n(n  +  1)  (3  m  —  n  -\- 1) 
6 
609.    Sch.     The  number  of  balls  in  an  incomplete  pile  is  the 
number  in  a  complete  pile  having  the  same  base,  diminished  by 
the  number  in  a  complete  pile  whose  base  would  be  the  next 
course  above  the  top  course  of  the  incomplete  pile. 

EXAMPLES  CXLVII 
Find  the  number  of  balls  in  each  of  the  following : 

1.  A  triangular  pile  of  15  courses. 

2.  A  triangular  pile  of  20  courses. 


INTERPOLATION  375 

3.  An  incomplete  triangular  pile  of  15  courses,  having  21  balls 
in  the  top  course. 

4.  An  incomplete  triangular  pile  whose  bottom  course  has 
15  balls  on  a  side,  and  whose  top  course  contains  28  balls. 

5.  A  square  pile  of  15  courses. 

6.  An  incomplete  square  pile  whose  bottom  course  has  20  balls 
on  a  side,  and  top  course  8  on  a  side. 

7.  A  rectangular  pile  whose  bottom  course  is  42  balls  by  20. 

8.  A  rectangular  pile  whose  top  row  contains  23  balls. 

9.  An  incomplete  rectangular  pile  whose  top  course  is  12  balls 
by  20,  and  whose  bottom  course  is  52  balls  in  length. 

INTERPOLATION 

610.  Interpolation  is  the  process  of  introducing  between  the 
terms  of  a  series  other  terms  which  conform  to  the  law  of  the 
series. 

The  most  extensive  use  of  interpolation  is  in  finding  inter- 
mediate terms  between  those  given  in  mathematical  tables,  and 
in  finding  right  ascensions,  declinations,  etc.,  for  other  times  than 
those  given  in  the  Nautical  Almanac. 

611.  The  Argument  is  the  variable  quantity  on  the  value  of 
which  the  magnitude  of  the  function  depends. 

Thus,  in  finding  from  the  table  on  pages  300  and  301  the 
logarithms  of  given  numbers,  the  given  numbers  constitute  the 
argument.  In  finding  from  the  Nautical  Almanac  the  moon's 
declinations  for  given  times,  the  given  times  constitute  the 
argument. 

When  the  changes  of  the  argument  are  proportional  to  the 
changes  of  the  function,  no  formula  is  needed  for  interpolating. 

612.  Prob.  To  interpolate  between  two  consecutive  terms  of  a 
series  a  term  that  shall  conform  to  the  law  of  the  series. 

Solution.  Let  p  be  the  distance,  in  intervals,  of  the  required 
term  t  from  the  first  term  a.     Then  p  is  an  improper  fraction  and 


376 


HIGHER   A  LG  Eli  HA 


is  one  less  than  the  number  of  terms.     Substituting  p  +  1  for  n 
in  the  formula  for  the  ?<th  term  of  a  series  (Art.  599),  we  have 

p(p  _  i)  p(p  _  i)  (*  _  2) 

=  a  +  p A  +        |0 — z  A  +  — nr^— 


1? 


|3 


A+- 


613.  The  value  of  t  is  more  closely  approximated  as  more 
orders  of  differences  are  used,  and  is  absolutely  correct  when  the 
terms  of  the  next  order  of  differences  are  0. 

For  facilitating  computation  the  values  of  the  coefficients  of 
A?  A  A?  an(l  A>  have  been  computed  for  every  hundredth  part 
of  an  interval  and  arranged  in  a  table.  See  Loomis's  Practical 
Astronomy,  page  393.    * 

614.  The  formula  of  Art.  612  is  for  increasing  values  of  the 
argument.  A  similar  formula  could  be  deduced  for  decreasing 
values  of  the  argument.  When  one  of  these  formulas  gives  a 
positive  error,  the  other  usually  gives  a  negative  error.  Bessel's 
Formula  is  obtained  by  taking  the  half  sum  of  these  two  for- 
mulas. Representing  the  argument  by  A,  and  the  function  by  F, 
we  have 


Argument     Function 


A-2 
A-l 
A 


-4  +  1 

A  +  2 
A  +  3 


F 
F' 


F"' 
Fiv 


1st  diff. 


A 

A' 

A"- 

A" 

Aiv 


2d  diff. 


A 


J^2 

A'" 


3d  diff. 


I), 


4th  diff. 


A 


A' 


ith  diff. 


IK 


For  interpolating  between  A  and  A  +  1,  F'  is  taken  as  the  first 
term,  making  p  a  proper  fraction ;  and  the  formula  becomes 


m 


+ 


p(p  +  l)(p-l)(p-2) 
I* 

15 


A-h 


INTERPOLATION 


377 


Bessel's  table  (see  Loomis's  Practical  Astronomy,  page  392) 
gives  the  coefficients  as  far  as  the  fifth  differences  for  every 
hundredth  part  of  an  interval. 

For  most  purposes  the  first  three  terms  of  the  above  formula, 
employing  only  first  and  second  differences,  are  sufficient.* 

EXAMPLES  CXLVm 

1.  An  eclipse  of  the  moon  occurs  only  at  the  time  of  opposition, 
i.e.,  when  the  sun  and  the  moon  are  on  opposite  sides  of  the  earth. 
For  the  eclipse  of  June  12,  1900,  the  Greenwich  mean  time  of 
opposition,  as  given  by  the  Nautical  Almanac,  is  15  h.  31  m.  30  s., 
and  the  right  ascensions  for  the  14th,  15th,  10th,  and  17th  hours 
are  as  given  below.  Find  the  right  ascension  of  the  moon  at  the 
time  of  opposition. 

Solution 


Argument 

Function 

1st  cliff. 

2d  diff. 

3d  diff. 

hr. 

hr.      min.        sec. 

June  12,  14 

17     19     56.00 

min.      sec. 

2     24.94 

sec. 

15 

17     22    20.94 

.10 

A 

16 

17     24    45.98 

2     25.14 

.10 

u 

17 

17     27     11.12 

In  this  case  F"  =  17  h.  22  m.  20.94  s.,  p  =  31  m.  30  s.  =  .525  h., 
A"  =  2  m.  25.04  s.  =  145.04  s.,  DJ  =  .1  s.,  and  D2"  =  .1  s.  Sub- 
stituting in  the  formula  of  Art.  614,  we  have* 

t  =  7  h.  22  m.  20.94  s.  +  .525  x  145.04  s.  +  .525  (-  .475)  x  .1  s. 
=  17  h.  22  m.  37.074  s. 

2.  The  Greenwich  mean  time  of  conjunction  at  the  time  of  the 
total  eclipse  of  the  sun  on  May  28, 1900,  as  given  by  the  Nautical 
Almanac,  is  2  hours  57  minutes  2.7  seconds,  and  the  right  ascen- 
sions of  the  sun  for  two  noons  before  and  two  after  are  4  hours 


*  For  a  fuller  discussion  of  the  subject  of  interpolation,  see  Loomis's 
Practical  Astronomy,  Chauvenet's  Practical  Astronomy,  Doolittle's  Practical 
Astronomy. 


378  HIGHER   ALGEBRA 

15  minutes  13.84  seconds,  4  hours  19  minutes  17.40  seconds,  4 
hours  23  minutes  21.43  seconds,  and  4  hours  27  minutes  25.90 
seconds,  respectively.  Find  the  mean  time  of  that  phase  of  the 
eclipse  which  occurs  at  the  instant  of  conjunction,  for  New 
Orleans,  which  is  6  hours  west  of  Greenwich. 

3.  The  right  ascensions  of  Jupiter  on  four  successive  days  at 
noon  being  10  hours  5  minutes  38.6  seconds,  10  hours  6  minutes 
18.86  seconds,  10  hours  6  minutes  59.41  seconds,  and  10  hours  7 
minutes  40.24  seconds,  respectively,  find  the  right  ascension  for 
midnight  of  the  second  day. 

4.  The  cube  roots  of  60,  62,  64,  and  66  being  3.91487,  3.95789, 
4,  and  4.04124,  respectively,  find  the  cube  root  of  63. 

5.  On  Oct.  29,  1900,  the  altitude  of  the  sun  when  on  the 
meridian  at  Minneapolis  was  31°  32'  17. "17.  The  south  declina- 
tions of  the  sun  at  Greenwich  apparent  noon  on  Oct.  28th,  29th, 
30th,  and  31st  were  13°  3' 49. "8,  13°23'53."2,  13°  43' 43. "9,  and 
14°  3'  21. "4,  respectively.  Find  the  latitude  of  Minneapolis, 
which  is  6  hours  12  minutes  56.8  seconds  west  of  Greenwich,  the 
latitude  being  the  complement  of  the  sum  of  the  meridian  altitude 
and  the  south  declination. 


CHAPTER   XXV 

PERMUTATIONS  AND   COMBINATIONS 

SECTION  I  — PERMUTATIONS 

615.  Permutations  are  the  different  orders  in  which  tilings,  taken 
the  same  number  at  a  time,  can  be  arranged. 

Thus,  the  permutations  of  the  letters  a,  b,  c,  taken  two  at  a  time, 
are  ab,  ba,  ac,  ca,  be,  cb,  and  taken  three  at  a  time,  abc,  acb,  bca, 
bac,  cab,  cba. 

616.  Theorem.  The  number  of  permutations  of  n  things  taken  r 
at  a  time  is 

„Pr  =  n(n-l)(n-2)(w-3)  •••  (n-r +  1). 

Dem.  Let  nP2,  nP3,  nP4,  •  •  •  nPr  be  the  number  of  permutations 
of  the  n  things,  according  as  2,  3,  4,  •••  r  things  are  taken  at  a 
time. 

Taken  two  at  a  time,  each  of  the  n  things  may  be  placed  in 
turn  before  each  of  the  remaining  n  —  1  things,  giving 

nP2=7l(7l-l). 

Taken  three  at  a  time,  each  of  the  n  things  may  be  placed  in 
turn  before  each  of  the  (n  —  1)  (n  —  2)  permutations  that  may  be 
formed  of  the  n  —  1  remaining  things  taken  two  at  a  time,  giving 

nP,=  n(n-l)(n-2). 

Taken  four  at  a  time,  ea.ch  of  the  n  things  may  be  placed  in 
turn  before  each  of  the  (n  —  1)  (n  —  2)  (n  —  3)  permutations  that 
may  be  formed  of  the  n  —  1  remaining  things  taken  three  at  a 
time,  giving         .  nA  =  ,l(n  _  1}  ()(  _  2)  („  _  3). 

The  number  subtracted  from  n  in  the  last  factor  is  seen  to  be  in 
each  case  1  less  than  the  number  of  things  in  each  permutation ; 
hence         ^  =  n  ()i  _  1}  (w  _  2)  (n  _  3)  ...  (b  _  r  +  X) 

379 


380  HIGHER  ALGEBRA 

617.  Cor.  The  number  of  permutations  of  n  things  taken  all  at  a 
time  is  nPn  —  [w. 

In  this  case  r  =  n. 

618.  Theorem.  The  number  of  permutations  of  n  things  taken 
all  at  a  time,  ivhen  p  of  one  kind  are  alike,  q  of  another  kind  alike, 

and  so  on,  is 

\n 

[p_  X  \q  X  etc. 

Dem.  Let  N  be  the  number  of  permutations  that  can  be 
formed. 

If  in  any  one  of  these  N  permutations  the  p  like  things  were 
replaced  by  p  unlike  things  (unlike  one  another  and  unlike  the 
other  n  —  p  things),  by  changing  the  order  of  these  p  unlike  things, 
leaving  the  other  n  —  p  things  unchanged,  this  single  permutation 
would  furnish  [p  permutations  (Art.  617).  The  same  change  in 
each  of  the  N  permutations  would  furnish  N  x  \p_  permutations. 

As  the  same  reasoning  applies  to  the  q  like  things  and  to  the 
other  sets  of  like  things,  the  replacing  of  the  different  sets  of 
like  things  by  unlike  things  would  furnish  N  x  \p_  X  \q  X  etc. 
permutations.  But  the  n  unlike  things  would  furnish  [n  permu- 
tations (Art.  617);  hence 

N  X  [p_  X  \q_  X  etc.  =  \n, 

"  [» 

whence  N  = , r== 

\jpx\qx  etc. 

619.  Theorem.  If  p  specified  things  are  required  to  occupy  speci- 
fied places,  the  number  of  permutations  is  the  same  as  of  n—p> 
things  taken  r  —  p  at  a  time. 

Dem.  The  only  permutations  possible  are  those  arising  from 
changes  of  the  n—p  things,  of  which  r—p  are  in  each  per- 
mutation. 

620.  Cor.  If  the  p  things  can  be  rearranged  among  themselves, 
the  number  of  permutations  is  \p  times  the  number  of  permutations 
of  n—p  things  taken  r—p  at  a  time. 

621.  Theorem.  The  number  of  permutations  ofn  things  taken  all 
at  a  time  in  a  circle  is  \n  —  1. 


PERMUTATIONS  381 

Dem.  Since  for  every  arrangement  the  tilings  may  all  be  shifted 
the  same  number  of  places  in  either  direction  around  the  circle, 
only  relative  positions,  and  not  actual  positions,  are  to  be  con- 
sidered. Hence,  if  any  one  of  the  things  remain  in  any  one  of  the 
positions,  all  possible  permutations  will  be  formed  by  permuting 
the  remaining  n  —  1  things  among  the  remaining  n  —  1  positions, 
giving  |  m  —  1  permutations  (Art.  617). 

622.  Cor.  If  each  order  of  arrangement  is  limited  to  one  direc- 
tion around  the  circle,  the  number  of  permutations  is  \  \n  —  l. 

It  is  to  be  noted  that  a  right-hand  arrangement  as  viewed  from 
one  side  of  the  circle  is  identical  with  the  corresponding  left-hand 
arrangement  as  viewed  from  the  other  side  of  the  circle. 

EXAMPLES  CXLIX       ' 

1.  In  how  many  different  orders  can  3  hats  be  hung  on  8 
hooks  ? 

2.  How  many  different  numbers  of  3  figures  each  can  be 
formed  from  the  digits  1,  2,  3,  4,  5  ? 

3.  How  many  different  numbers  of  2  figures  each  can  be  formed 
from  the  9  digits  ?  How  many  of  3  figures  each  ?  Of  4  figures 
each?  Of  5  figures  each?  Of  6  figures  each?  Of  7  figures 
each  ?     Of  8  figures  each  ?     Of  9  figures  each  ? 

4.  In  how  many  different  orders  can  6  persons  be  seated  at  a 
dinner  table  ? 

5.  In  how  many  different  orders  can  3  persons  occupy  7  fixed 
seats  ? 

6.  In  how  many  different  orders  can  4  players  use  6  billiard 
cues? 

7.  In  how  many  different  ways  can  4  gentlemen  select  from  7 
ladies  partners  for  the  waltz  ? 

8.  In  how  many  different  orders  can  a  single  platoon  of  8 
soldiers  be  arranged  in  line  ? 

9.  How  many  numbers  between  1000  and  10000  can  be  formed 
by  use  of  the  digits  1,  2,  3,  4,  5,  6  ? 


382  HIGHER  ALGEBRA 

10.  In  how  many  different  orders  can  the  letters  of  the  word 
Algebra  be  arranged?  In  how  many  the  letters  of  the  word 
Mathematics  f 

11.  How  many  different  signals  may  be  made  with  4  different 
colors,  taken  any  number  at  a  time  ? 

Sug.     The  number  =  4Pi  4-  4P2  +  4 Pz  +  4P4. 

12.  In  how  many  different  orders  can  8  boys,  any  number  at  a 
time,  enter  a  room? 

13.  In  how  many  different  orders  can  12  members  of  a  minstrel 
troupe  arrange  themselves  in  line  on  the  stage,  the  same  two 
always  acting  as  end  men  ? 

See  Arts.  619  and  620. 

14.  A  shelf  contains  20  books,  of  which  4  are  single  volumes 
and  the  others  are  in  sets  of  8,  5,  and  3  volumes  respectively. 
Find  the  number  of  ways  in  which  the  books  can  be  arranged  on 
the  shelf,  (a)  when  the  volumes  of  each  set  remain  in  the  order  of 
their  number,  (b)  when  the  volumes  of  each  set  are  together,  but 
in  any  order. 

15.  Find  the  number  of  permutations  of  the  factors  of  a2b3c4. 

16.  In  how  many  different  orders  can  a  football  eleven  play,  if 
the  full-back,  the  half-back,  the  quarter-back,  and  the  center-rush 
always  play  in  the  same  positions  ? 

17.  Four  of  the  crew  of  an  eight-oared  boat  have  trained  to  row 
only  on  the  stroke  side  and  four  to  row  only  on  the  bow  side.  In 
how  many  ways  can  the  captain  arrange  his  crew,  (a)  when  the 
stroke  (the  rower  who  sets  the  stroke)  is  any  -one  of  the  four  on 
the  stroke  side,  (6)  when  the  stroke  is  always  the  same  man  ? 

18.  Either  A  or  B  of  a  baseball  nine  must  pitch,  either  C  or  D 
must  catch,  while  E,  F,  and  G  must  play  on  the  bases.  In  how 
many  ways  can  the  captain  play  his  team  ? 

19.  In  how  many  different  orders  can  6  persons  be  seated  at  a 
round  table  ? 

20.  In  how  many  different  orders  can  7  children  stand  in  a 
ring? 


com  r>  ix  Ai 'ioxs  383 

21.  In  how  many  different  orders  ean  a  host  and  7  guests  sit  at 
a  round  table,  the  host  always  having  the  guest  highest  in  rank 
on  his  right,  and  the  guest  next  in  rank  on  his  left  ? 

22.  In  how  many  different  orders  can  a  party  of  5  ladies  and 
5  gentlemen  sit  at  a  round  table,  the  ladies  and  gentlemen  sitting 
at  alternate  places  ? 

Explain  why  this  is  [4  x  |JS. 

23.  In  how  many  different  orders  can  10  beads  be  strung  for  a 
rosary  ring  ? 

See  Art.  622. 

24.  In  how  many  different  orders  can  5  like  pearls,  6  like 
rubies,  and  7  like  diamonds  be  strung  for  a  bracelet  ? 

25.  If  the  number  of  permutations  of  G  things  is  360,  how 
many  are  taken  at  a  time  ? 

26.  If  nJP2  =  30,  what  is  n? 

27.  If  nPn  =  40,320,  find  n. 

28.  If  nP6  =  10  x  nP,,  find  n. 

29.  If  nP5  =  12  x  nP3,  find  n. 

30.  If  nPr  =  990,  what  is  r  ? 

31.  If  2nP:i  =  100  x  nP2,  what  is  n  ? 

SECTION  II  — COMBINATIONS 

623.  Combinations  are  the  different  groups  into  which  things, 
taken  the  same  number  at  a  time,  without  reference  to  the  order 
of  arrangement,  can  be  formed. 

Thus,  the  combinations  of  the  letters  a,  b,  c,  d,  taken  two  at  a 
time,  are  ab,  ac,  ad,  be,  bd,  cd;  and  taken  three  at  a  time,  abc,  abd, 
acd,  bed. 

While  ab  and  ba  are  different  permutations,  they  are  the  same 
combination. 

624.  Theorem.  TTie  number  of  combinations  of  n  things  taken  r 
at  a  time  is 

_n(n--l)(n-2)(n-3)-(n-r  +  l) 

[r 


384  HIGHER  ALGEBRA 

Dem.  Let  nGr  be  the  number  of  combinations  of  the  n  things 
taken  r  at  a  time. 

By  Art.  617  each  combination  of  r  things  can  have  \r  permuta- 
tions. Hence  the  number  of  combinations  is  the  number  of 
permutations  divided  by  [r;  that  is  (Art.  616), 

_  ?i(n-l)(n-2)(7i-3)  —  (n  -  r  +  1) 

625.  Theorem.  The  number  of  combinations  of  n  things  taken 
r  at  a  time  is  the  same  as  the  number  of  combinatioyis  of  n  tilings 
taken  n  —  r  at  a  time. 

Dem.  For  each  combination  containing  r  things  there  is  left  a 
combination  containing  the  remaining  n  —  r  things ;  hence, 

fl  —    C 

626.  Sen.  In  numerical  applications  the  formula  of  Art.  625 
often  involves  much  less  labor  than  that  of  Art.  624. 

Thus,  in  determining  the  number  of  combinations  of  15  things  taken  12  at 
a  time,  the  formula  of  Art.  624  gives 

c     _  15  .  14  •  13  •  12  .  11  .  10  .  9  •  8  .  7  • 6  •  5  •  4 
15    12~       2.3.4.5.6.7.8.9.10.11.12      ' 

while  that  of  Art.  625  gives  simply 

„        15-14-13 
15°3=       2-3      ' 

627.  Theorem.  If  p  specified  things  are  to  be  included  in  each 
combination,  the  number  of  combinations  is  the  same  as  of  n  —  p 
things  taken  r  —  p  at  a  time. 

Dem.  The  only  combinations  possible  are  those  arising  from 
the  combinations  of  the  n—p  things,  of  which  r  —  p  are  in  each 
combination. 

628.  Theorem.  The  total  number  of  combinations  of  n  things 
taken  any  number  at  a  time  is  2n  —  1. 

Dem.  Each  thing  may  be  treated  in  two  ways,  as  it  may  be 
included  or  excluded.     Either  way  of  treating  any  one  of  the 


COMBINATIONS  385 

things  may  be  combined  with  either  way  of  treating  each  of  the 
other  things,  giving  as  the  number  of  such  combinations 

2  x  2  x  2  x  •  •  •  to  n  factors  =  2n. 

But  this  includes  the  case  in  which  all  the  things  are  excluded. 
Hence  the  total  number  of  combinations  is  2n  —  1. 

629.  Cor.        .Q  +  nC,  +  jCt  -f «-  nCr  =  2"  -  1. 

630.  Theorem.  The  total  number  of  combinations  of  n  things 
taken  any  number  at  a  time/  when  p  of  one  kind  are  alike,  q  of 
another  kind  alike,  r  of  another  kind  alike,  and  so  on,  is 

(P+i)(9  +  i)0-  +  i)... -i. 

Dem.  The  p  like  things  may  be  treated  in  p  + 1  ways,  as  all 
may  be  excluded,  or  1,  2,  3,  •••  ov  p  included. 

Similarly  the  q  like  things  may  be  treated  in  q  + 1  ways,  the 
r  like  things  in  r  -+- 1  ways,  and  so  on.  Hence  the  total ,  number 
of  ways  of  treating  all  the  things  is  (p  -f-  1)  (q  +  1)  (r  +  1)  •••. 

But  this  includes  the  case  in  which  all  the  things  are  excluded. 
Hence  the  total  number  of  combinations  is 

0>  +  i)(<,  +  i)(>-  +  i)--i. 

631.  Cor.  TJie  total  number  of  combinations  of  n  things  taken 
any  number  at  a  time,  when  p  of  one  kind  are  alike,  q  of  another 
kind  alike,  r  of  another  kind  alike,  and  so  on,  ivhile  the  remaining  t 
things  are  unlike,  is 

O>  +  i)(7  +  l)0'  +  i)-"(2')-l. 

This  follows  from  Arts.  630  and  628. 

632.  Theorem.  The  number  of  combinations  of  p  +  q  +  r  +  ••• 
things  ivith  p  things  in  each  combination  of  one  set,  q  in  each  of 
another,  r  in  each  of  another,  etc.,  is 

\p  +  q  +  r+  ♦•- 


[p  X  [q  X  \r_  X 
dowxey's  alg. — 25 


386  HIGHER  ALGEBRA 

Dem.     Multiplying  numerator  and  denominator  of  the  formula 
of  Art.  624  by  \n—_r,  it  becomes 

c  =n(n-  1)Q -  2)  ...  (n  -  r  +  1) (n-r)  —  3  •  2  •  1  =         |» 

|r(*— r)  •••3.2.1  [r  x|w-/ 


Since  for  each  combination  of  p  things  there  is  left  a  combina- 
tion of  n—p  things  (Art.  625),  the  number  of  combinations  of 
p  +  q  things,  with  p  things  in  each  combination  of  one  set  and  q 
in  each  of  another,  is,  by  the  above  formula, 

\p  +  q 


\px\q_ 

If  there  are  p  +  q  -\-r  things,  the  number  of  combinations,  with 
p  things  in  each  combination  of  one  set  and  q  +  r  in  each  of 
another,  is 

\p  +  q  +  r 

\px\q  +  r 

But  the  number  of  combinations  of  p  +  q  things,  with  p  things 
in  each  combination  of  one  set  and  q  in  each  of  another,  is 

text1 

hence  the  whole  number  is 

\p  +  q  +  r      [g  +  r       \p  +  g  +  r 
x 


\p  X  \q  +  r     \qx\r_     [p  X  \q  X  \r 

As  the  same  reasoning  applies  to  any  number  of  sets  of  com- 
binations, the  formula  is  as  stated  in  the  theorem. 

633.    Cor.     If  there  are  m  sets  of  combinations,  and  p=q=r=---, 
the  above  formula  becomes 

\mp  \mp 

either    ~~~     or 


(\p)m  \m  x  (\p)m' 

according  as  the  combinations  of  the  different  sets  are  treated  as  dis- 
tinct or  identical. 


COMBINATIONS  387 

EXAMPLES  CL 

1.  How  many  different  couples  can  be  selected  from  6  people  ? 

2.  How  many  different  sums  of  3  figures  each  can  be  formed 
from  the  digits  1,  2,  3,  4,  5  ? 

3.  How  many  different  products  of  2  factors  each  can  be  formed 
from  the  nine  digits  ?  How  many  of  3  factors  each  ?  Of  4  factors 
each?  Of  5  factors  each?  Of  6  factors  each?  Of  7  factors 
each  ?     Of  8  factors  each  ?     Of  9  factors  each  ? 

4.  How  many  sets  of  fours  can  be  formed,  at  different  times, 
from  6  soldiers  ? 

5.  How  many  different  committees  of  5  each  can  be  formed 
out  of  a  corporation  of  12  members  ? 

6.  From  a  company  of  50  soldiers,  how  many  pickets  of  6  men 
each  can  be  formed  ? 

7.  In  how  many  ways  can  4  vacancies  be  filled  from  10 
applicants  ? 

8.  How  many  span  can  be  formed  from  10  horses  ? 

9.  A  druggist  accepts  an  offer  of  $  5  for  as  many  glasses  of 
soda  water  as  can  be  flavored  with  any  two  of  his  20  sirups.  Does 
he  gain  or  lose  by  the  transaction,  and  how  much,  the  price  being 
5  cents  per  glass  ? 

10.  In  how  many  ways  can  a  base-ball  nine  be  selected  from 
15  players,  the  pitcher  and  the  catcher  being  always  the  same 
men? 

11.  How  many  different  amounts  can  be  weighed  with  5  weights 
of  1  pound,  2  pounds,  4  pounds,  8  pounds,  16  pounds,  respectively  ? 

12.  How  many  different  products  can  be  formed  from  5  differ- 
ent factors,  taken  any  number  at  a  time  ? 

Query.     Why  must  5  of  the  combinations  be  rejected  ? 

13.  In  how  many  different  ways  can  12  stops  of  an  organ,  any 
number  at  a  time,  be  opened  ?  . 

14.  How  many  different  sums  can  be  formed  with  a  three-cent 
piece,  a  five-cent  piece,  a  dime,  a  quarter  dollar,  a  half  dollar,  and 
a  dollar  ?  Would  the  answer  be  the  same  with  any  six  pieces  of 
money  of  different  values  ? 


388  HIGHER  ALGEBRA 

15.  In  how  many  ways  can  a  selection  from  5  rubies,  4  dia- 
monds, and  3  emeralds,  taking  at  least  one  of  each  kind,  be  made  ? 

16.  In  how  many  ways  can  a  selection  from  5  pears  and  6 
apples,  taking  at  least  one  of  each  kind,  be  made  ? 

17.  With  5  dimes  and  5  half  dimes,  how  many  different  amounts 
could  be  put  into  a  contribution  box  ? 

18.  With  3  weights  of  one  denomination,  5  of  another,  and  4 
of  another,  how  many  different  amounts  can  be  weighed  ? 

19.  From  5  apples,  4  apricots,  an  orange,  a  pear,  a  peach,  and 
a  banana,  how  many  choices  of  fruit  may  be  made  ? 

20.  From  a  committee  of  10,  how  many  subcommittees  of  2,  3, 
and  5  members  respectively  can  be  appointed  ? 

21.  From  12  soldiers,  how  many  different  scouting  parties  of 
2,  4,  and  6,  respectively,  could  be  formed  ? 

22.  In  how  many  different  ways  can  the  52  cards  of  a  pack  be 
divided  equally  (a)  among  4  players,  (b)  into  4  piles  on  the  table  ? 

Sug.  When  divided  among  the  players,  the  mere  exchange  of  what  they 
hold  would  give  different  hands  to  the  respective  players ;  but  when  divided 
into  piles  on  the  table,  the  exchange  of  positions  would  not  give  different 
piles.    See  Art.  633. 

23.  In  how  many  different  pairs  can  a  railroad  section  boss 
distribute  his  10  men  for  work  along  the  track  ? 

Query.     In  what  way  does  this  differ  from  example  1  ? 


ANSWERS 


Examples  I 

2.  2  m2  +  2  mn  +  2  n\  3.  6  a6  -  0  a2x  +  7  ax2  +  ax3.  4.  x4. 
5.  15  a2cx2  +  2  a26x2  +  7  wixty*.  6.  ab2  -  2  a26  +  a3  -  6«c  -  4  ae2  -  c3. 
7.    a2  -  xy.            8.    a263  -f  x2y.            9.    4  a2  -  62  +  a26  +  a62  +  2  63  -  3  a3. 

10.    6w2  +  2»2  +  2fm2«  +  li»i»2.      11.    4x?-x£  +  5x2  +  5x-*/-a6-x3-3. 

12.    5c2  -  Vx  +  2V2. 

Examples  II 

1.    (a  +  6c -36  -  2c  +  4)x.  2.    2  6x +(c  -  5  6)y. 

3.  (a  +  4)x  +  (2  6 -3)y+(c-4)z.  4.  (a  +  »)(*  +  */)  +  (6  -  n)(x  -  y). 
5.    2  a.        6.    (2  +  4 a)  Vx  -  y.        7.    fv/o^x.         8-    JL  +     ~m  +  Sb. 

.  vx       y 


9.   5  a Vm2  -  x2.         10.    (a  4-  6  +  c)  Vx2  -  **, 

Examples  III 

2.    x3  +  x2-2x-8.       3.    4xy.       4.    6x+2x3.       5.    3x3  -  13x2  +  9x-  3. 

6.    2a3  -  2a2  +  4a.  7.    4a2.  8.    x4  +  5 x3y  -  2 x2?/2  -  0 xy\ 

9.    3a  +  36  +  3a'  +  3c.  10.    GVX  +  2VX3.  11.   4x^yi 

12.    3(x  +  y)+6(x-*/).  13.    2  6VT372.  14.    |  vaTx2. 


15.   2(a-6  +  c)Vx2  +  y2.  16.    (a  -  6)  Vx  +  y  +  (a  +  6)  Vx  -  y. 

17.   3x2  +  7x-8.         18.    2x3-4x2y-  5x*/2  +  3?/3.         19.    3a -66  + 4c. 
20.    6  vx  +  y  +  (a  -  6)  Vx  4-  y  +  (c  -  a*)  (x  -  y). 

Examples  IV 
1.    3x2-3x-5.         2.    6a -46 -2.         3.    3a +  2.        4.    4x2-4x  +  5. 

5.  6m +  2.  8.    -(3-a)x3-(c-26  -  4)x2-(2d  -  4)x. 

6.  2z.  9.    -  (c  -  «2)x3  -  (6  -  a  +  5)x2. 

7.  x  +  y  +  z.  10.    -(6-a)x3-(6  +2c)x2-(6+  c  +  d)x. 

389 


390  ANSWERS 

Examples  V 

5.   20«565.       6.    72ab~1c3.       7.    -42x3^.       8.    160x^"i      9.  30a667c3. 

10.    72aWy5-     11.    -42xt*/2.     12.  -  216  a%c~ h     13.  -180z3™+»+y»+4«+4. 

Examples  VI 

2.  6  a*  -  23  a3?)  +  41  a2&2  -  42  a&3  4"  18  b*.      3.  3 x4  -  7  x3  -  4x2  +  16x  -  8. 

4.  8x5  -  6x4  -  25x3  +  13x2  +  18 x  -  8.  5.  x°>  +  y\  6.  x4  4  afy8  4  */4. 
7.  ra6  -  am2)^2  +  »«  +p6.  8.  a2"1  -  0"+*  4  am+2  -  am+1  +  an+1  -  a3. 
9.    8  a3  4  27  63  -  c3  4  18  abc.       10.   a3  +  63  -(-  c3  -  3  a&c. 

12.  2x74  10  x6-  9x5  +  21x4-22x3-llx2  +  24x-  12. 

13.  G  x6  -  3  xby  4  xV2  -  1 1  xy5  +  15  ^6. 

14.  ?^8  -  10  m6  4  5  mfi  4  25  m4  -  43  m3  -  12  m2  +  44  m  -  40. 

15.  x7  +  x6  +  x5  +  16x4  -  24x2  +  41  x  -  15. 

16.  3x9  +  10  x7  -  4  x5  -  9xl  +  16x3  +  6x2  -  12. 

17.  8  x4*>  4  6  x^yi  +  3  x2^2?  4  22  xPy*q  4  6  y*. 

Examples  VII 

3.  9x4  +  3x3-2x2  +  6x-4.  4.   x5  -  5x4  4  10x3  -  10x2  4-  5x  -  1. 

5.  6  «4  -  10  a3x  -  22  «2x2  +  46  ax3  -  20  x4.  6.    81  x4  -  yK 
7.   6  as9  +  10  x5  -  38  x4  +  11  x3  +  46  x2  -  17  x  -  15.  8.    m5  4  4  mrc4. 
9.    x">  -  50  x4  4  769x2-3600.                                                     10.    x8  4  x4  +  1. 

11.  x5  -  5  ax4  +  10  a2x3  -  13  a3x2  +  13  a4x  -  6  a5. 

12.  x6  4  3  xhy  -  3  x4y2  -  11  x3y3  +  6  xV  +  12  xif  -  8  jfi. 

13.  2x7  -  11  x6  +  14x5  +  17  x4  -  50 x3  +  23x2  +  38x  -  30. 

14.  x7  -  7  x6  +  21  x5  -  17  x4  -  25  x3  +  0  x2  -  2  x  -  4. 

Examples  VIII 

2.  x4  -  x3.y  -  13  xV  -  40  x?/3  4  25  y4. 

3.  2x5  4  7x*y  +  3x3y2  +  8x2?/3  -  5x?/4  4  300 y*>. 

4.  4  m6  4  12  mbn  -  24  m4n2  4-  6  m3n3  -  72  ra%4  4  36  ran5  -  20  n6. 

6.  5  x7  -  15  xhj  -  6  x5?/2  4  18  x4*/3  -  4  x3*/4  4  12  x2y°  +  7  xy*  -  21  y7. 

7.  x5  4  3  x4*/  4  28  x3?/2  4  3  x2?/3  4  12  x?/4  -  63  ?/5. 

8.  a5  4  8  «4&  -  43  «362  4  67  a2?;3  4  72  a&4  -  144  65. 
10.  2  x5  4  13  x4  4  19  x3  4  25  x2  4  31  x  4  30. 


ANSWERS  391 

11.  4  x5- 10^4-  10x3  +  7x24-3x  +  4. 

12.  x6  +  4  x5  -  16  x4  4-  37  x3  +  8  x2  -  50  x  -  56. 

13.  x64-6x6-8x3-48x2  +  5x  +  30.  14.    Sx?  -  27  x4  -f  7x  -  63. 
15.  5x'i  +  60xs-5x-60.                 16.   x7  -  10  x6  +  3x*  +  30  x3  +  6  x  -  30. 

Examples  IX 

2.  x5-3x*-  6x3  +  12x2-40x  +  96.  3.   x4  -  9x3  -f  16 x2  -  9x  -  35. 

4.  x4  +  4  x3  -  19  x2  -  46  x  +  120. 

5.  2  x6  +  3  x5  -  25  x4  -  13  x3  +  11  x2  -  26  x  +  168. 

6.  x5  -  9X4  +  x3  +  105  x2  -  74  x  -  168. 

7.  3  x6  -  7  x5  -  51  x4  +  41  x3  4-  96x2  -  142x  +  60. 

8.  9  &  -  18  x5  -  193  x4  +  206  x3  +  444  x2  -  88  x  -  160. 

9.  x7  -  2  x6  -  10  x5  +  28  x4  +  5  x3  -  74x2  +  76  x  -  24. 


5.  x2  +  4  x  -  60. 

8.  m2+  19  m  +  48. 

11.  y2  +  80|/-900. 
15.  x4-5x2-84. 
20.  m'W  -  10  ?nx3  -  39. 

23.  a6x«  -f  20  a3x3  -  69. 
27.  9  z6  -  18  z8  -  72.  28.    25  z4  -  100  z2  4-  75.      30.    x2  -  16  xy  4-  60  y*. 
31.  x2  4-  16  xy  4-  48  ?/2.        32.  x2  -  16  xy  4-  63  y2.        33.    ^4-4  xhj  -  32  y2. 
34.  4z44-8z2y-45y2.       35.    ^  _  4  &tf  _  32  ^6.       36.    ^  -  4  z3*/2  -  77  y*. 
37.  9x64-15xV-  14  y4.  39.    x2  -(8y  -  3  z)x  -  24  yz. 
40.  x*  +  (11  y -4  z)x* -Uyz.  41.   9X6  4  3(7  y-  6  z)x*  -  35  yz. 

42.    25  a4  4-  5(6  wi8  -  3  n^a2  -  18  m%2. 

Examples  XI 

1.    1  4-  2  x  +  x2.      2.  x2  4-  4  x  4-  4.      3.  x4  4-  2  x2y  4-  y2.      6.  x2w  4-  4  xrt  4-  4. 

7.   £*,+£  8.    ^-l*-*-.  9.   xt4-4  +  ^-. 

2/2  *2  16x1  Syl     36^ 

12.  a%  -  a//3  4-  }a*  66.     13.  x*n  -  2  x^yn  4-  y2n.     16.  (16  +  56  xn  4-  49x2«)x2. 

17.   49x3-42xV^  +  9v"5.  18.    i^4:2  4-^-  19.    x4-4y2. 

9  o2n  \alm 

20.    9  ?>i4  -  25  nc>.        21.    1  -  $  a2.  22.  25  x6  -  9  y4*2.  23.  4  xf  -  9  y. 

24.  (81  ax  -  l)ax. 


Examples  X 

6. 

X2  _  4  x  _  96. 

7. 

m2  4-  16  m  -  80. 

9. 

x2- 18x  + 77. 

10. 

y*-l8y-88. 

12. 

x2  +  25  x  -  1250. 

14. 

x*  _  14  x2  4-  48. 

16. 

x6  +  9  x3  -  70. 

17. 

zs  _  25  z4  -  150. 

21. 

ro4z8  4-  9  m2z4  -  90. 

22. 

a6x4  -  16  «8x2  - 

25. 

4  z'2  -  4  x  -  63. 

26. 

16  x4  -  12  x2  -  8 

392  ANSWERS 

Examples  XII 

1.  a2  4-  b2  f  c2  -  2  ab  -  2  ac  +  2  5c.    2.  a4  +  £>4  +  c2  -  2a262  +  2a2c  -  262c. 

3.  m2  4-  n2  +  p2  +  q2  —  2  row  —  2  mp  +  2  rog  4  2  wp  —  2  nq  -  2pq. 

4.  x4  +  4  y2   +  «2  +  9  v2  +  4  x2y  -  2  x2!^-  6  x2v  -  4  yit  -  12  yv  +  6  uv. 

5.  x6  4-  2  x5  4-  3  x4  4-  4  x3  4-  3  x2  +  2  x  4- 1. 

7.  -x4  -  2  x3*/  4-  3  x2y2  -  2  x*/3  4-  ?/4. 

8.  x6  4-  2  x5y  4-  3  xV  +  4  x3y3  +  3  x2?/4  +  2  x?/5  4-  y6. 

9.  9x44-25y24-  16^4-4w2-30x2?/4-24x%3-12x2w-40^34-20yM-16  2;3M. 
10.    a8  4-  4  66  4-  9c4  +  16 d2  -  4  a463  -  6  a4c2  4-  8 a*d  +  12  63c2  -  16  bH  -  24  c2d. 

Examples  XIII 

2.  x2  -  6  xy  4-  9  y2  -  2rdz2.  3.  x4  4-  x2y2  +  */4.  4.  4  x2  -  4  xz  4-  *2  -  16  y2. 
5.  16  x8  +  23  xly4  4-  9  y*.  7.  x4  -  2  x3  +  x2  -  81.  8.  x4  -  x2  4-  14  x  -  49. 
9.   9x4  4-  14  x2  4-  25.  10.  a2  +  lab  +  4  b2  -  9  c2  -  6  cd  -  d2. 

Examples  XIV 
1.    2  a2.       2.    5x2.        3.    4x%"2.        4.    3xm-».        5.   6xw+n.         6.    a2xi 

m—n  , 

7.    a^x-"-.         8.    (ab)Sm.  9.   6x"5.         10.    ro2x3.          11.    a2(a  -  x)2. 

12.   4ro(a2-3x)5.  13.    33(x2  -  y2).                   14.   a2(2x  4-  4x3)". 

15    <WT           16    3«3d3x5  17      4x"+w 

.'.  &3y3#               '      5fc2c   '  '    7ym+nzP 

Examples  XV 

1.   2  ab2  4-  5 a36  -  4  ab.  2.    4x2y  -  12y2  +  9.         3.   -  3x2  4- 4ax  -  a2. 

4.    5  a2  -  3  b2  -  4  c2..       5.    3  xm  +  2  x2"»  -  4  x3"1.       6.  2  a™  &s~2n  -  3  «"*+2n  b*. 

7.    7  a2064  -  &8  +  13  aV2b*.      8.    x*  4-  3  xV  _  2.      9.   a1-2"-  ^-"-a  +  a^". 

10.    2{x-y)  -3(x-*/)24-4(x-?/)3. 

Examples  XVI 

2:    5x2-4x4-3.  3.    2  x2  -  7  x  -  8.  4.    2  a%  4-  3  a2fc2  -  a&3  4-  4  64. 

Examples  XVII 

1.  5  a2  -6  ax  -  2  x2.  2.  x2  -  5  «x  4-  4  a2.  3.  2  ?/4  -  8  y3  4-  12  y2  -  8  ?/  +  2. 
4.    x2  -  3  X  4-  5.         5.    2  a3&  4-  3  a2b2  -  a&3.  6.    a3  4-  3  a2x  +  3  ax2  4-  x3. 

7.    9x4-6?/4-3z.     8.   2y3-4x«/24-x2y4-3x3.     9.  3 x3 4-  x2y  - 4 xy2 4- 2 y3. 


ANSWERS  393 

10.  l  -  x  +  xs  -  x4  +  2  x5.  11.   4  x2  +  6  x  +  9.  12.    -  x2  -  2  x  -  4. 

13.  2a2-a6  +  262.  fc    ?    2      5  2    2     7 

16.  x  +  y.  2x2  +  5xy  +  7y2 

17.  a4  -  a86  +  a262  -  a&3  +  64.  •  18.  x4  +  Zxhj  +  8x2y2  -  8y*. 
19.  |a2_^a_j.  20.  6x-iy-|.  21.  ax2-2x  +  6.  23.  3x2+2x  + 1. 
24.  x2-3x  +  7.  25.  2  a*  -  6  a26  +  18  ab'2  -  27  63.  26.  2x3-x  +  l. 
27.  2  a3  +  4  a'2  +  8  a  +  16.                    28.    a  +  b.                      29.    mw  +  anan. 

30.  2a3-3a2H  a&2  +  4  63. 

31.  2  x4  —  3  x3y  +  4  x2y3  —  5  xy3  +  6  y4,  with  remainder  4  y6. 

Examples  XVIII 

2.    7  x2  +  11  xy  +  20  y2.  3.    x3  +  G  x2y  -  xy2  -  30  y3. 

4.  2  x3  +  3  x2y  -  3  xy2  +  4  y3.      5.    12  x3  +  11  x2y2  -  «/3,  with  remainder  3  y4. 

6.  x4  +  x*y  -  5  x2y2  -  3  xy3  +  3  y\       7.   3x4  +  Sxhj  -  5x2?/2  -  24xy3  -  24 y*. 

8.  3x4  +  2  x*y  +  3 x2y2  -  4  xy3  +  2  y*. 

9.  4  x5  -  4  xV  +  3  x3y2  +  x2*/3  +  6  xy4  +  5  yb. 

10.    x5  -  8  x*y  +  S  x3?/2  +  4  x2?/3  +  2  y5.        12.    x3  +  8  x2  +  6x  +  2. 
13.    3  x4  -  9  x3  +  27  x2  +  4  x  -  15.  14.    8  x2  -  40  x  +  100. 

15.   5x3-6x2  +  7.  16.    x4  +  5x3+5x2+5x  +  5. 

17.    X4  -  x3  +  3x2  +  6  x  -  3.  18.    2x3  4-  5  x2  +  5x. 

19.    x5  +  5x4  +  3x3-4x2-2x  +  5.  20.    x4  -  2x3  +  x2  +  2  x  +  3. 

22.   x3  +  3  x2  +  9  x  +  27.  23.    x4  -  x3y  +  x2y2  -  xf  +  y*. 

24.    x5  +  3  x*y  +  9  x3?/2  +  27  x2y3  +  81  xy4  +  243  y*. 

Examples  XIX 
2.    x2-3x+2.     3.X+7.     4.  x3-3x2  +  2x-3.     5.    1.     7.x2-6x-2. 

Examples  XXI 

2.   x2+3x  +  2.  3.    x2  +  3x  +  2.  4.  x4  +  x2  -  4x  -  4. 

5.  x4  +  4x3  +  6x2  +  4x  +  l.      6.   x3-2x2  +  3x-5.    7.   3x3  +  4x2  +  5x  +  7. 

Examples  XXII  to  XXVIII 

The  student  would  derive  little  benefit  from  these  examples  if  the  answers 
were  given. 

Examples  XXIX 

2.    12a2&2.      3.    3x2?/2.      4.    3  w2(x  -  y)\      5.   x2y2(2x  +  y).      6.  x  +  2. 

7.  x  -  2.     .    8.    x  -  1.       9.   x  +  1.      10.  2x  +  1.      11.  a  +  b.     12.    x2-  x. 
13.   x2  +  2  xy.        14.   x  +  0.         15.  x  -  3.        16.  x2  +  4  x  +  3.      17.    x  +  2. 


394  ANSWERS 

18.  x2-2x  +  4.  19.    x-1.  20.    x2-3x-4.  21.    x2  +  3x-4. 

22.  x2  -  6  x  +  8.       23.    x2  -  x  +  1.       24.    4(x2  -  2  x?/  +  */2).       25.  a  -  2  6. 

26.  a2-3a&-462.  27.  3x-4.  29.   x3  -  6x2  +  llx  -  6. 

30.  x3  +  5x2  -2x-24.  31.    x4  +  4x3  -  2x2  -  12x  +  9. 

32.  3x4  +  4x3-7x2-4x  +  4.  33.   2x3  -  3x2  +  4x  -  5. 

Examples  XXX 

2.   x2  +  x  +  1.  3.    3  x  -  4.  4.   x2  +  xy  +  3  y2. 

5.   x4  —  x3y  +  x2y2  —  xy3  +  y4. 

Examples  XXXI 

2.    3x-3.  3.    7x  +  l.  4.    2x2-3x  +  3.  5.   2x2-3x+l. 

Examples  XXXII 
1.    x  +  6.  2.    x2-8x+15.  3.    2(x  +  2/). 

Examples  XXXIII 
2.   x2-4.  3.    (*  +  y}*(x -  y)*.  4.   x3  +  7x2- 9x-63. 

5.    x3-  9z2-x  +  105.  6.   x4-  10x3-39x2  +  460x-700. 

7.    x4  +  9  x3  +  11  x2  -  81  x  -  180.  8.    x3  -  2  x2  -  5  x  +  6. 

9.    (6x2-  7x-5)(x-4)(2x  +  4).     10.    a5  +  2a4  -  10a3  -20a2  +  9a  +  18. 
11.    x*-16y*.  12.    x4+ 5x3  +  5x2  -  5x-6.  13.    a4  -  1. 

14.  x5  +  7  x4  -  10  x3  -  70  x2  +  9  x  +  63. 

15.  (2x-3*/  +  4)(3x  +  2?/-5)(4x-y  +  3). 

Examples  XXXIV 

1     ^.         2    2ac"  3    ?•**&,  4    *  5         1  6    3_a 

7  c  '36  llc2n  "a*  '   x2  +  y2' 

7.   *±1  8.    ^L3.  9.    _J_.  10.    ErJ. 

x  —  1  x  +  5  x  +  1  x  +  3 

12     a2  -  ax  +  x2  ]3    x2  +  3x  +  9  H    x-5 

a  +  x  x+1  '   x  —  3 

16.    *-**.  17.    EH*.  18.    a  +  6  ~  c.  19. 

1—  4a  y  a  +  6  +  c 

^    2x  +  1  21     a  +  b  +  c  +  g  22    x  +  3| 

'    2  x  +  3  "    6  —  a  —  c  —  d  "    x  —  3 

24    2g  +  6        25        x~3      .       26.    x-ii-       27.    2  g  +  3- 

7x-5  x(3x-7)  x-5  3x-l  4x  -  3 


f 

46 

11    1  + 

X 

(1- 

X)2 

15.  l~y- 

a  +  6  —  c 

-a" 

a  —  6  +  c 

-d 

23.    5 

x  ■ 

-3 
-5 

28.    *x 

-5 

AN  S  WEBS  395 


Examples  XXXV 


5.    2x-4-— .  3.   3x2-2x  +  2+     ' 


3x  x-2 

4.   2x24-2xy4-  10  y2  +  — £~  5.    m  +  ft 


x  —  3  y  wi  +  w 

6.   3x-2-  5a;~  1.  7.    16  x4-24x3  +  36x2- 54x  +  81. 

4  x2  +  3 

8.    2  x  -  1  +      X~6 —  9.    4x24-6x-2 


x1  -  x  -  1  2  x2  4-  x  -  3 

10.  5x-14  4--^L-  11.  x  +  y. 

x  +  2 


Examples  XXXVI 

63                   18  x2 
a*                3x-4 

3.         4"X.           4. 

a  —  x 

X3 
X 

-y8 

4-y 

5.    ai+» 
a  —  b 

6     (a  +  64c)(Hc 
2  6c 

-a)            7    x2  4- 3x4-1. 
x4-  5 

8. 

a(a  4-  b) 
a  —  b 

Examples  XXXVII 

x(a-b)     y(a  +  b) 

*      .        o          ** 

6 

4  4-  4x 

aa  _  7>2        a2  _  fta      0a  _  62  6(x  _  x2)     6(1  -  x2)         6(1  -  x2) 

3  x*  -  x3y  4-  x2y2  -  gy8    x4  4-  x2y2         x3 

x4  -  y4  x4  —  y4      x4  —  y4 

4  x(a2-62))    y(q-fe)2)    g(q  4-  ft)2 
(a2  -  62)2'    (a2  -  62)2'    (a2  -  62)2' 

6    2x(x4-y))       3y        4g(x2-xy  +  y2) 

x3  4-  y3      x3  4-  y3  x3  4-  y3 

6  (x  -  y)2    x(x  -  y)  x2 

(x  -  y)3'    (x  -  y)3'    (x  -  y)8' 

7  X4-2  2(x-2)  3(x  4-  3) 

(x  4- 3) (x2  -  4)'    (x4-3)(x2-4)'    (x4-3)(x2-4) 

8  (x  4-  2)2  X4-2  1 

x3  4- 6 x2  4- 12x4-  4'   x3  4- 6 x2  4- 12x4-4'    x3  4-  Ox2  4-  12 x  4-  4* 

9  (x-2)2(x4-l)2    (x  4- 2)2(x  -  l)2    (x  4- 2)2(x  4- l)2t  (x-2)2(x-l)2 

X4_5x2  +  4     '        X*-5s2+4     '        x4_53'2  +  4     '        x*-5«a  +  4      , 

Examples  XXXVIII 

2       *«  3     -2—  4       2b*    .  5    _-J 6.   5£±i. 

x2  -  y2  x  -  x3  6»  -  a2  2  x  -  x2  x4  -  1 

7    2x  -  a  g  x  -  5  9    x  +  4  1Q         1 

x  +  a"  '    (x-3)(x-4)'  x-4  x3  - 1 


396 


n. 


x  +  2 


(!-*}(*  T3) 


12. 


ANSWERS 
1 


1  +  2x 


13.    0. 


15.    0. 


16. 


3x2  +  2x+  13 
x3  +  27 


17. 


14.         3(x2  +  l) 
x4  +  x2  +  1 


o     9qgft«y 
4x3 


(a2  -  63)5 


11.   J 

16. 
20. 


12. 

x  —  3  y 
x  +  3y 

x2  +  6  x  +  8 
x2  +  8  x  +  15* 


3. 

a 

x  +  y 

8. 

3x  +  1 

x  —  5 

1  - 

y 

Examples  XXXIX 

a     z  +  4 


9.    a2  +  1  4-  ^ 


10. 


-£-.         6.  ^-c. 

x-2  a3 

2  x4  +  6  x3  -  36  x2 


17. 


13.    1. 


14. 


18. 


(x-6)(x2-6) 
1 


x(x2  -  ?/2) 
x  +  3 


1    6<*ft2r 

12  xy2' 
6    x(x  +  7) 

10      2(a  -  6)2 

3  62(a  +  6) 

14.    *«  +  *        15. 

a  +  2x 


j    ad/ft  —  bcfh 
bdeh  +  bdfg 

fi    a  +  3 


2(x2  +  y2) 
21.    1. 

Examples  XL 

x2  —  3y  „     x  —  ?/ 

x  +  4  y  3  «x 

7     x2  +  xy  +  y2  g 

x2  —  xy  +  y2 

11.    a2  -  2  ac  +  c2  -  b2. 
1.         16.    1.         17.    1. 

Examples  XLI 


x-1 


1     an  —  bm 

ab 

x  —  a 

x  +  a 

12     ^x  ~ 

1 

2x- 

o 

18.    *  +  8 

7. 


2.   a-b.  3 

ag  +  ft2  +  C2 
a262  +  a2c2  +  62c2' 


4. 


15. 


19. 


x  +  y 

3  x  +  1 

x  +  3  " 


5.   3(a  +  &). 

x(x  +  2y) 

x+  1 


13. 


x  +  5 


2a 

2(w  -  ri) 
(ro  +  w)2 ' 


5. 


x-y 


i   ^  +  x  +  z 

yz  +  1 
4.    1. 


Examples  XLII 

bdf+be  +  cf' 
5.    (a3-53)2. 


x2  +  x  -  1 


3(x+l) 


ANSWERS  397 


Examples  XLIII 

«     xyc4  2    ab*  ~  aS.  3    *l2? 

a2fc3'                                    '    &•  +  a*b  '    (x  -  y)3(x8  +  y8) 

4       afrgg(q  -  6)                   5    ^(y2  -  x)2  g        (1  -  x2)*> 

y8(a2  -  aft  +  b2)'                '    y(y  +  x2)8 '  '   x"»+2p(x2  -  y)* 

Examples  XL IV 

1.    VtfW.         2.    ViV$.         3.    — * 4.    b&  5. 


\/(ax)8  3  7</x8^ 

>/7  -      3Va"7  y/s«  Q     ^Iv'x^"' 


3 

a*x3 


io.  *y.      a.  b*v».     i2.  is*      is.  «y. 

yt  *" 


*+t  *+t 


15.    «P*T.         16.    £±$4         17.    (a2+2&)*(x  +  */)l       18.    M+3&. 


Examples  XLV 


1.    a6*)4.  2.    a46~4c.         3.    a*b  M.  4.    36xV2-  5.   ^2^3. 

6    6W.  7     ^  8     *te.  9    3W  ^ 

11     bahfi  12.    ***  13.    J^L  14     **  15     8a3c^ 


3  6*„  5^  *  66 

16.    16x4^.  17.    |*  18.    *3T. 


Vy 


Examples  XL VI 

2.  x4  +  4  xfy  +  6  x2y2  +  4  x?/3  +  y4. 

3.  «5  -  5  a46  +  10  a362  -  10  a268  +  5  ab*  -  65. 

4.  8&-12x*y  +  Gxy2-y3.  5.    1  -  4x2  +  Ox4  -  4x«  +  x8. 

6.  a6  +  6  «56  +  15  a4b*  +  20  a363  +  15  a2b*  +  6  abb  +  &c. 

7.  x8  -  4  x«y2  +  G  xV  -  4  x2^  +  y8. 

8.  x2  +  8  x$y  +  24  xy2  +  32  x^3  +  10  y*. 

9.  1  +  10x*  +  40  x4  +  80  x6  +  80x8  +  32x10. 

10    a2  ,  8  A2     24  aft4     32  A8     16  68< 

X  X2  X8  X* 


398 


ANSWERS 


11.  x6  -  3x5  +  -^x4  -  | x3  +  ^x2  -  T\x  +  &. 

12.  a2  +  8  aM  +  28  A  +  56  aM  +  70  ab2  +  56  a*^  +  28  ah*  +  8  aM  +  64 

13        1         3«V      15  oV     20  flV      60  a»y"     06a*y«  ■  6i    1g  1<t 
'    64x12      8x10         4x8  x«  x4  x2  "  * 

14.    x4  -  4  xBy  +  6  x2*/2  -  4  x?/3  •+  ?/4  +  4  x*z  -  12  x2?/z  +  12  xy2z  -  4  -fz  +  0  x2z2 
-  12  xyz2  +  6  */222  +  4  xz3  -  4  */23  +  z4. 


15.    8x2  -  4V(x  +  1)8(«  -  1)-  4V(x  +  l)(x  -  l)3  -  4. 


16.    32 x6  +  6x5Vx2  -  1  -  48 x4  +  20x3(x2  -  1)*  +  18x2  +  6x(x2-  1)*-1. 

Examples  XL VII 

2.  8  x6  -  36  x5  +  114  x4  -  207  x3  +  285  x2  -  225  x  +  125. 

3.  216  x6  -  432  xby  +  504  x4^2  -  64  x3*/3  +  168  x2y*  -  48  xy5  +  8  if. 

4.  64 x9  -  96  x8  +  192  x7  +  88 x6  -  96 x5  +  366x4+147x3-15x2+225x+125. 

5.  x12  -  6  x11  +  21  x10  -  47  x9  +  81  x8  -  108  x7  +  126  x6  -  117  x5 

+  99  x4  -  61  x3  +  42  x2  -  12  x  +  8. 

Examples  XL VIII 

5.    5  x2  +  2  y  -  z\  9.    6  a3  -  2  b2  +  3  c  -  5  & 

Examples  XLIX 

3.    6x3-3x2-x.  10.    4  x4  -  4  x2y2  -  7  y4. 

16.    4x3-5x2  +  2x-3.  20.    11  x3  -  Sx2y  +  bxy2  +  9y\ 

28.   4  x6  -  6  x4  +  2  x3  -  3  x2  +  9.  32.    3x6-5x5-4x2  +  2x-5. 

33.  6  x6  -  4  x5  -  2  x4  +  x3  +  5  x2  -  3  x  -  7. 

34.  7  x7  +  5x6  -  3  x5  -  x4  +  8  x3  +  4  x2  -  2  x  +  6. 

Examples  LII 
3.   2x4-3x2-l.       6.    4x3-2x2  +  3x  +  5.      9.   3  x5-2  x4+x3+4z2-3x--5. 

Examples  LIV 

3.  5V3. 

6.  15  VS. 

9.  Za2b2VTa. 

12.  21  a&x2V5~6x. 

15.  9xyVSy. 

18.  20  amb»+3x/2a™. 


2.    4V2. 
5.    14  V2. 

8.    iabVWb. 
11.    24x^2V3. 
14.    2a6c2vT&2. 
17.    12  am+1&2nV3a^. 


4.   6V2. 
7.   24  y/E. 

10.    10  6c3V5a&c. 
13.    60a2rax2V2m. 
16.    YobcVbabtf. 
19.    -  21  a262c3^4&. 


ANSWERS 

39! 

20. 

22  xy2VSx. 
a\/a  +  b. 
x(x  +  y)  V2. 

21. 
24. 
27. 

36  a6-2c2V5l;. 

22. 
25. 
28. 

20  a£r2c2 #7^. 

23. 

3xV2x-3y. 

(x2  -  xy2)  V3x. 

2  a2&V5  a  -  8  a26. 

26. 

2av/3a-46. 

Examples  LV 
V7.      3.   VlS.      4.  aVl56.       5.    #7.      6.    VB.       7.  &VlO~a.      8.   VS. 


14. 

1. 
2. 

4. 
10. 

14. 


6. 

9. 
11. 
13. 
15. 
18. 
21. 


a\/bb.       10.    V4.        11.    -a</9b*c-       12.    V12  am+*.       13.   -  V2  a. 
y/UaW2.      15.  3v/3aP.      16.   V3a6.     17.  4Vx  +  2y.      18.  2Vx-3y. 

Examples  LVI 


\/9a^,   V49  xfy*,   Vi2hy,  V4  x2  -  12  x*/2  +  9  J/4. 
#64  x6^,   #125  <rx9,   #-27  a-663c6,   #216  x3*/6*"9. 
VIO.         5.    #2.         6.    y/\.         7.    #8~I.         8.    -W 

>x  +  y 


Vx(a  +  6). 


11.    #a-6. 


12. 


3.   VT«. 

9.    #2x. 
13.    (a2-62)*. 


Jw  +  w.  15     (4a;a_4y2)l  i6.    (a*  -  a*)!.  17.    Vx2  -  1. 

*  in  —  n 

Examples  LVII 
3.    #32,   #8.  4.    #32,   #9.  5.    #256~^,   WW¥. 


#8,   #9. 


a-"1,    V64m.  7. 

#8T«*,  #8~P,   y/mV2. 


IT 


'ft*.  8.    #a6,   #M,   #c3. 

10.    #8l,  #86,  #fl. 


V04a6,    v8a3,    v4a2 


12-    #a3+3a26+3a&2+&3,  #a2-2a&  +  62. 
14.    3#125a3x^,  2v/x4y2,  4#12^3. 
5#100x«,  7#256xV,  6#343  xV5.  16.    #7.  17.    #3. 

#16.  19.    ML  20.    #x*  when  x^l,    #x»whenx<l. 

Vx^  when  x>l,    v/x3  when  x<l. 


Examples  LVIII 
f#3.  5.    ^V5x.  6.    \VQ.  7.    &V42. 

ay/x8  +  2/ 


8Vx2-2 


«vr  —  y8 
b(x  -  y)  ' 


10. 


b(x  +  y) 


x2-2 
11.  f#l5.        12.  2V5I.       13.  4>/55x. 


14.    f#9.       15.    f#36.  16.    i-^iex2.       17.   *#54.       18 


4  a  3, 


7  6x 


4x 


19. 
24. 
27. 


9   6, 


$#48.        20.    3#768.        21.    bV$a\        22.    ~  #18xy .        23.    4#4x. 
^—-^#13^.        25.    (2  +  x)#(2-x)8.         26.    (x2  -  xy  +  y2)  #xT7- 

(x2-t,2)#(x2+y2)*.       29.   3+V2.      30.   (^+^)2.       31.    14  +  5v^. 

x-y  11 


400 


ANSWERS 


I.     5(VT+ »*+»). 


33.    2  x2  -  2  xVx2  +  1  +  1. 


35.    x  +  Vx2 


36.    Vx2  +  a.      37.   x  +  vx2-l. 


38 


34.    x  +  Vx2  -  1 . 
Va2+62x2-a 
6x 


40.    2(2+V2-V6).  41     (2V2+V8-V5)(2V0-3) 

y  30 

Examples  LIX 

2.    v/9  +  v/6  +  ^i.  3.    4-2^4  +  2^2. 

4.  \/32  +  4\/5  +  2\/2v/5  +  2V5  +  V2v/25  +  \/55. 

5.  xzVx  +  x2  Vx2  Vy»  +  x2y  Vy2  +  x?/2  Vx  Vy  +  ?/3  v'x2  Vy  +  y4  V^. 


Examples  LX 

39V2.     3.  10V3  +  21V5.     4.  59V3.     5.  72V7.     6.    8a/6.     7.    26VTL 
7V3.  9.   SVl.  10.    V0V2.  11.   9  VI.  12.    9abV2~aV2. 

27x\/2x.  14.    2x^/Cx2.  15.    70a2V2.  16.    (9  +  4x)v/2x2. 

13x(3  +  2x)V2.         19.    16V6.         20.    2VI5.        21.    ^2-Va.        22.    0. 
VS.       24.    VS.       25.    ^VlO.        26.    f\/U.       27.    20 VS.      28.   4\/9. 


13. 
17. 
23. 


29.    17a6V2a2-362 
33.    88-4  V0. 


30.    Z>abVb  +  2a. 


31. 


2  a2 


1. 

8. 
14. 
20. 
25. 
31. 
36. 
42. 
45. 


Examples  LXI 
C.        2.    7V3.        3.    10.        4.    3.        5.   6x. 
xy.        9.    48.        10.    14V6.        11.    30  V2. 
13V21.  15.    77a/C.  16.   91 V2. 

20^108.         21.    12  VS.         22.    2aVab. 
Vim).        27.    \Zl35.  28.    3^32. 


6.    15V30. 
12.    5V42. 
17.   iV5. 
23.    ^27. 
29.    \/2. 


2_a 
x 


7.    72  V2. 

13.    35V91. 

18.    iVlO. 

24.    3x\/3x. 

30.    l^v^P. 


V(l+x)8. 


J2.  V (a2 -62)2.  34.  2  +  7V3.  35.  12x-6  +  16V2x. 
3+V15  +  VC.  37.  7.  38.  33.  39.  3.  40.  a3.  41.  4-2VI0. 
2+2VT5.  43.  -48  +  54VG  +  12Vl0+60\/l5.  44.  9a2+3a&+4  62. 
18  a  -x.         46.    x  +  y.         47.    3^0-12^3-^180+12. 


Examples  LXII 

3.    4.  4.    bVS.  5.    2.  6.    \V2.  7.    2^2^.  8.    2a\/2, 

9.    Vx2  -xy  +  y2.       10.    Vx2-Sy.       11.    fv35.       12.    VT5.       13.    VS6. 

14.   -VStf.       15.    Vl.       16.    V2.       17.    $£       18.    -  VWab.       19.    ^V6. 
a  b 


ANSWERS 


401 


20. 


28. 


1.        21.    y/E.         22.    V5-3V3.         23.    8-8V3  +  V5.         24.    V7. 
V3  -  1^243  +  £\/27.  26.    2  +  VB.  27.    -^  +  f  VlO. 

Va  +  V&+Vc.  29.   ^  +  f*2-  SO.   |V6. 


x4  +  aW. 
a4 


1.   75  a;. 


Examples  LXIII 
2.    24  z.        3.    16x\/25^.        4. 


54  «62. 


8a 
25 


V5a. 


6. 

9Vl4a. 

10. 

aby/Tb*. 

14. 

12-2V35. 

1. 

6. 
11. 

3. 
8. 

4. 

8. 

12. 

3. 

9. 

14. 

18. 

22, 
30. 
37. 


7.    abVWb.  8.   32aVa26.        9.   2a3&2V2a3&2. 

11.    5y/2xy>  12.    -SabK  13.    2xy/y. 

15.    15  +  6V6.       16.    9-6V2.      17.    3Vl8  -  3VI2  -  1. 
18.    v^  +  3v36-f  3v48  +  2. 

Examples  LXIV 
y/ab.        2.    6v/5a^.        3.    -3V40&2.        4.   a?by/2aF*.        5.    2v/7ax. 
2\/xp.  7.    Va.  8.    aVl.  9.    y/43xy*.  10.    x(x  -  y)n. 

y/1  -  5  a2.  12.    2Vo6. 

Examples  LXV 
1  +  VS.        4.    1  +  y/2.        5.    V5  -  V2.         6.    V6  -  V2.         7.   5  f  V6. 

y/l-  V3.  10.    VT0  +  2V2.  11.    V8-V3.  12.    2  +  3V5. 

Examples  LXVI 

8a  +  15&V^T.            5.    78V^T.            6.    28V^T.  7.   5xV^~I. 

2V^T.           9.    18V3~V^T.            10.    14V^1.           11.  5aV2~V-I. 

10+8V^3.       13.    yfcX.       14.    6+V^a.       15.   2V^2.  16.    vCl. 

Examples  LXVII 
-xy.  4.  -20Vl5.  5.  180.  6.-4 
-2(14+V2).  10.    4(11  -V^T). 

l&f/Sy/^lyY^l.       15.    16\^73\^T\^ 

256  y/l. 

23.    2. 


19.    972V2V-1.  20.    - 

25.    V3L        26,  27.    5.        28. 


7.    -42V3. 

8.    -  72. 

11.    2. 

12.    153. 

[.        16.    -50. 

17.    -24. 

128V2V^T. 

21.    2304 

»+*£*.        29. 

7  +  2Vtf 

11 


W-i. 


31. 


19 -6  VlO.       34.    12.       35.    §V3. 


5 
V^l. 


3V-T.       38.    7V4.       39.     v^V^T.       41.    V-l.       42.   §±|£3. 

21 


DOWNEY'S    ALG. 


20 


402 


ANSWERS 


43. 


i*-x  +  2a- 


a?  +  x 
48.  3-4V^l. 
52.   6-V^i. 


57. 


A 


44.    1. 


45    2(<*2  ~  *0 
a'2  +  62 


8. 


49.   4  +  3\/^l. 
54.    7-4VC1. 

6  +  2\/5x-2v/2x-6 


47.    2  +  3' 


1. 


50.    5-2V-1. 
55.   6  +  8.\Cl. 


3. 

-10. 

4.    -55. 

10. 

1. 

11.    -1. 

17. 

3. 

,«    ac 
18.    -• 

1. 

81. 

2.    c2-2  6c. 

7. 

1. 

8.   6.        9.    6. 

Examples  LXVIII 
5.   23$.  6.   |.  7.    7. 

12.    4.  13.    ^.  14.   i 


19.   0. 


21.    0. 


8.    4.  9.    4. 

15.    4|.  16.    2. 

22.    -f. 


Examples  LXIX 
3.    16.  4.    5. 


5.   4(a-l). 


10.    3. 


11. 


2aVb 

b  4  r 


12. 


i4.  («±^),        i5.,{i-(|^)n. 

17.  ±<*-l*       18-  f       19.  v«2-(^y 


81 

a 

16. 


6.   8. 
13.   4. 

4  m2 


(m  + 1)2 

(a  -  I)2 
2a 


21. 

22.   *£. 

23.    1.            24.   4. 
Examples  LXX 

3. 

35.       4.             an? 
nq  —  mq  — 

•       5.    112|.       6.    an  +  hm.       7.    ii.        8.    175. 
np                                an  +  bn 

9. 

T\.         10.    40.         11 

,    30.         12.    ftf         13-   6-        14-    77-        15-    40- 

16. 

1.                7.    $1200. 

18.    n(c~b),    n(a-c\              19    40)  ga 
a  —  b          a  —  b 

20. 

6,  15,  oo.          21.    5712.          22.    105.          23.    50.          24.    A,  32  ;  B,  25. 

26. 

f  3  hr.  16T4T  m. 
t  3  hr.  32T8r  m. 

27.    4  hr.  41T5T  m.               28.    4  hr.,  54  mi.,  60  mi. 

29. 

43.         30.    142,857. 

31.      225    .         32.    $1.         33.    30.         34.    300. 
m  -f  n 

35. 

15,  90.             36.    10. 

37.    8.            38.    10.            39.    140,  60,  45,  80. 

40. 

50,  $1350,  $1200. 

41.    45,  30. 
Examples  LXXI 

1. 

10,  7.         2.    10,  3. 

3.    4,  1.         4.    2,  2.          5.    2,  -  3.          6.    12,  9. 

7. 

1,  2.          8.    -  3,  5. 

9.    -  2,  -  4.           10.    -  2,  1.            11.    6,  -  10. 

ANSWERS                                              403 

12. 

4,  -  3.        13.    1,  -  2. 

14.    -  8,  5.         15.    18,  12.         16.    -  2,  -  ft. 

17. 

be'  —  b'c     a'c  -  acf 
a'b  —  ab'    a'b  —  ab' 

18.    36,       ?.       19.    a,«.      20.    «+26,  a"26. 
2          2                                          2     '      2 

21. 

4,    -0.          22.    -6, 
a  +  6,  — i-r 

-2.         23.    -i    1         24.    -\   -A         25.    !,    J. 

a    6                 6        a                 6    a 

26. 

a  +  6 

Examples  LXXII 

4.  18f,  31£.         5.    5,  60.         6.    30,  90.  7.   *£-         8.    3,  9.  9.    24 

10.    15,  22*.  11.    24,  30.  12.    bm  +  a,    an  +  b .  13.    48,  1G 

mn  —  1     mn  —  1 

14.    30,  20.                  15.    42,  11,  7.                  16.    50,  75.  17.    4£,  7$ 

18    pm  +  qn  -  qmn    pmn  -  qn  -  pm                19    g^  2Q  2Q    ^  g 
mn  —  m  —  n    '      mn  —  m  —  n 

21.  lO^,  17.             22.   24,  32.             23.  10,  24.             24.  $10,000,  $7200 

25.  100,  12.        26.    $3000,  4%.        27.    Ill,  9|.        28.   a(b  ~  c) ,  P(c  ~  a) 

b  —  a         b  —  a 

29.    50,  30.  30.    18,  15.  31.    48,  80;  $63,  $33.  33.    5%,  8%, 

34.    n(r~P\   <P  ~  g).  35.    11,  7. 

r  —  q  r  —  q 

Examples  LXXIII 

1.    1,  2,  3.         2.   2,  3,  4.         3.    3,  2,  1.         4.   2,  5,  -  1.         5.   j£,  -  J,  J. 

6.    25,  55,  65.  7.   3,  4,  7,  1.  8.    G,  3,  4,  5.  9.    -  5,  4,  -  3,  -  2. 

10.    2,  1,  1,  2.  11.    I,  2,  1.  12.   |,  -  J,  ft.  13.    3f,  2|,  2^- 

14.    £,$,*.        15.    2  a,  2  6,  2  c.        16.    |,  f,  f.        17.    f,  |,  2.        18.    1,2,3. 

19.   3$,  2§,  2f.  20.    -A&L,    _JL«£.,    -JL^.  21.    4,  5,  6. 

6+c        a+c        a  +  6 

22.  2,  1,  4,  3.       23.   1,  -  2,  4,  -  3.        24.   3,  2,  -  5,  1.        25.    1,  2,  3,  4,  5. 

26.  2,  1,   -2,  3,   -1,  4.  27.    2.  1,  3,  4.  28.    4,   -2,  1,   -3. 
29.    1,3,  5,  7.      30."  3,  4,  -  G,  5.      31.    4,  6,  8,  10,  12.      32.   5,  -3,  7,  -2,  4. 

Examples  LXXIV 
1.    426.  2.    $2,  20cts.,  lOcts.  3.    37,45,52.  4.   480,400,560. 

5.  3  7G0,  200,  200.  6.    $  1000,  $-  500,  $0.  7.    120,114,110. 
8.    1  to  2,  6  min.        9.   21,  17,  23,  15.         10.    18&,  34?,  23/r,  80. 

Examples  LXXV 
4.   The  1st.        7.   Less.        9.   The  1st.        11.   The  1st.        12.   Between  34 
and  37.      13.  Between  15  and  20.      14.  Between  $  12  and  $  14.      15.   15,  12. 
16.    7£.  17.    5.  18.    20  or  5. 


404  Answers 

Examples  LXXVI 

1.    z-5.  2.    -^~  3.    x  +  y.  4.    K«LzJ&  5.    5^8 

x-2  *  2(a  +  &)  a:_l 

6.    £.        7.    11.  8.    T\V  9.   f.  10.    Greater.  11.    Greater. 

12.    Less.  13.    15,  21.  14.    14,  35.  15.    7.  16.    5:8. 

17.       „    m2"2 •  18.     4:1.  19.   3  :  2.  20.    120,  160,  200. 

m2  -f  mn  +  w2 

21.    42,  12.         22.    8:9.         23.    3  : 4.         24.    242,  300,  358.         25.    16,  20. 
26.    B,  11  :  10.         28.    (a)  72,  (b)  84,  (c)  75,  (d)  77. 


Examples  LXXVII 
1.    f.        3.    f        4.    xy  -  —  •         5.    3.         6.    3,  12.         13. 


xy  a_&_c_l_d 

18.       -   ~ac —       20.  200,  300.     21.  8,  6.     22.  2  or  lOf.    23.  13,  26,  or  39. 

24.    Volte"      °25.    -!£-,   -!£_.       li'-'jL  +  Jtt   JL-.±1       27.78,66. 

8-o    *4«  2  m     2  a    2  m     2  a 

28.    .0089.       29.    &        31.    8:45  a.m.        32.    10.        33.    6.        34.   45,30. 


35.    100. 


Examples  LXXVIII 


2.   zxy.       3.   xk2*.        6.   x  =  8y.        7.  x  =  — .       8.   z  =  — .       10.   24. 

11.  5.  12.  9.  13.  \.  14.  4.  15.  6.  17.  48.  18.  579. 
19.    67£.        20.    18.        21.   47.        22.    250.        23.    26|.       25.    ysr-fs  +  ii. 

2  Sit 

26.   x  =  1  +  2  y  +  3  y2.  27.   s  =  $/**«  28.    r(- V.  29.   27,  74. 

30.    .00611.  **' 

Examples  LXXIX 

1.  58,590.        2.    83,903.        3.    -58,  -828.        4.    -3,0.        5.    Z  =  116, 
£=1404.        6.  a  =  l,  £=540.        7.   d  =  6,  £  =  473.       8.    n  =  12,  £=582. 

9.  a  =  32,    Z  =  84.  10.    d  =  -  5,    Z  =  -  95.  11.  n  =  18,    d  =  & 

12.  a  =  | ,  d  =  -  TV  13.  103,  133,  163.  14.  d  =  9.  15.  -  46. 
16.  f.  17.  576.  18.  1512.  19.  78th  and  79th.  20.  (2  t  -  l)a,  at2. 
21.  14,475.        22.    10  sec.,  4f  min.        23.    19,600  ft.        24.  11. 

Examples  LXXX 

2.  2048,  4095.  3.  19,683,  29,524.  4.  16,384,  21,845f£.  5.  3072,  2049. 
6.  a  =  5,  I  =  640.  7.  a  =  4,  £  =  39,364.  8.  I  =  -  ^,  £  =  -  $ 1. 
9.  a  =  l,  £=511.   10.  r=-4,  £=1638.   11.  r=±3,  £=2186  or  -1094. 

12.  n  =  5,  £  =  121.  13.  a  =  5,  «  =  6.  14.  68,  272,  1088.  15.  r  =  3. 
16.21.  17.13,120.  18.3072.  19.  ±  2.  20.2.  21.6,24,96,384,1536. 


ANSWERS  405 

22.   2.         23.    |.         24.    f.         25.    ft*.         26.    ft.         27.    tff        28.    \. 

29.    150.  30.    (a)  $127,093,291,416.30;    (6)  $25,418,658,283.26. 

(c)  $169,457.72. 

Examples  LXXXI 

1.    TiV  2.    TV,*V  3.    28.  4.    16|,21,28.  5.    *,*.*• 

Examples  LXXXII 

±5.  3.    ±  1.  4.    ±  1.  5.    ±  3.  6.    ±  Vl9 

8.    ±|.       9.  ±  v'2  aft  -  b2.       10.   ±  |  a/3.       11.   ±  a/3. 


1. 

±3. 

2. 

7. 

6 
-  1 

2. 

±  — 

2 

a/4  aft  -  P 

Examples  LXXXIII 

1.    6,  15.      2.  mV*  ,  wV*  ,  PV*         .      3.  6,  15. 

Vra2  +  w2  +  pa     Vm2  +  n2  +  .p2     a/mi2  +  n2  +  p2 

4.    12  ft.,  18  ft.     5.  4,16.     6.  40,36.     7.  4550.     8.  3.     9.  5.57  +  ,  18.12-. 

10.    161124.3  + .  11.       1  _  from  the  weaker  light. 

a/7±V17 
12.   _E^L  from  the  weaker  bell.         13.  5.  14. 


1  ±  a/3  a/2  n  —  m     a/2  n  —  m 

15.  20,  10.        16.    150.        17.   48. 

Examples  LXXXIV 

3.    5,  1.       4.   -1,  -5.       5.    11,  -1.       6.    -1,  -9.      7.    11,  1.      8.  8,  -6. 

9.    8,   -  2.  10.    6,  2.         11.    10,   -  6.         12.    8,   -  1.  13.    32,  -  2. 

14.   14,  -  2.        15.    15,  -  3.         16.    11,  -  3.        17.   4,   -  \.        18.    2,  -  1. 

19.    £   *L         20.    2,  -tt.         21.    *,    -*.         22.   «±h         23.    U*   |. 

ac    ac  a        b  c  4      2 

Examples  LXXXV 

3.    12,  36.       4.    14,  10.      5.    6.       6.    20,  3.       7.    12.       8.    70,  80.  9.    40. 

10.    7.     11.  10,10.|.     12.  3.     13.   18,  $20.     14.   15.      15.  576.      16.  12  or  8. 

17.    9,  36.        18.    1  hr.  48  min.,  2  hr.  15  min.        19.    3  or  4$.        20.  15. 

Examples  LXXXVI 

3.    5,  1.         4.  8,   -  3.         5.    3,  3.         6.    7,   -  7.         7.    0,  7.         8.    6,  6. 

9.   z2-2x  =  8.  10.   z2-10x  =  -21.  11.    x2  +  3 x  =  40. 

12.    x2  -  15  x  =  -  54.  13.    x2  =  3.  14.    x2  -  4  a;  =  1.  15.   6,  4. 

16.  6,2.         17.    2^2  =  9r/.         18.    ja-I. 


406  ANSWERS 

Examples  LXXXVII 
1.    3,  0.       2.    5.    (Reject  -  2).       3.    6,  7.        4.    1,  8,  -  5.        5.    -  1,  ±  4. 
6.    1,  JT-       7.  0,  4(a+  b).       8.  3,  2.        9.  4,  \.        10.  8,  -  §.       11.  4,  }. 
12.    ±V3.         13.    5,3.        14.    ±  |V3.         15.0,4,12.         16.    ±5. 

Examples  LXXXVIII 

1.  ±  4.     2.  3.     3.  36.     4.    ±  729.     5.   ±  125.     6.  27.     7.  J.     8.  f.     9.  64. 

Examples  LXXXIX 

2.  2.  3.  7,  -  1.  4.  ±  5.  5.  ±  2.  6.  -  3.  7.  ±  2. 
8.    K5±V5).        9.    -|(l±vl3).         10.    1,-5. 

Examples  XC 

2.    ±  2,   ±  1.  3.    ±  V5,   ±  1.  4.    ±  3,   ±  V^T.  5.    ±  3,   ±  4. 

6.  ^2,  -2.       7.  4,  v^iO.       8.  16,  ifffi.       9.  243,  (-28)1       10.  8,  iff. 

11.    ±|V30,  ±1.      12.  64,  (-33)1      13.   — ,  — 14.    v^,   V^VS*. 

15.    16,  TV         16.   2,  i.  V^     ^=6 

Examples  XCI 
2.    5,  -2|.         3.    1,  1.         4.    2,  -6|.        5.    9,  -12.         6.    ±5,  ±  3V2. 

7.  3,  -\.  8.  1,0,-3,-4.  9.  4,69.  10.  8,  -  1,  |(7  ±  V53). 
11.  2,  1,  |(7  ±  V33).  12.  2,  -  |,  K3  ±V505).  13.  3,  -.1,  1  ±  |V5L 
14.    2,   -|,  K-17±V305).  15.    1,  K-3±V5).  16.    5,  -2. 

17.    3,   -31,   i(- 1*^^251).  18.    K1±V5)-  19-    f(l±Vfy 

Examples  XCIV 

1    (3,-ff;       2/±Vf;       3    {2;     4(l,-lt;     5    (5,10;  f  a,  a ; 

*  \_4,jy^.  l2Wf-  *-2.        '1 2,  -8§.  I  10, 5.  \b,b. 

Examples  XCV 

f2,-5;  1.P.-1;  3    f-2,0;  4    U  ±  *V33  ;_ 

*  14,  -3.  '   10,  -f.  '12,  1.  '\_i_tiV33. 

Examples  XCVI 
.     fl,  -1|,8,  -2;  2    f±6; 

'  1 3,  —  3f,  —  3,  2.  '   I  ±  3. 

3    (3,  -2£,  Kl±v/105);  4    fl,  -  1$,  &(-  1  ±  V409); 

'   11.  -Jt,  Kl±Vl06).  "   I  2,  -2|,TV(-1±V409). 


ANSWERS  407 

Examples  XCVII 


•     fiVifVfj  8f±2,  ±$>/lO;  3    f±3,T8; 

'I  ±2,  ±iV2.  "li^TlVlO.  '\±5. 

4   /±1.  ±¥VE?;  gf±2,  ±^V3_;  r±i0|±ttV=47 


Examples  XCVIII 


3,-5 
5,6. 


6    f  ±  fV2l ;        fi    f  2,  10,  -  2,  -  10  ;         -    f  10,  -  5  ;         g    f  ±  3,  ±  2 
'l±j>/21.  '11,-3,-1,3.  '1 5,  -10.  \±2j±3. 

Examples  XCIX 
4,2,  £(_13±V377);  3    /  3,  lf8±v^j 


2,  4,  i(-13TV377).  '  1 1,  3,  2  =F  V^. 

4    (0,12,6;  5    (4,2,  K~7±V^35); 

'  1 0,6,  12.  '   12,  4,  K-7TV-35). 

Examples  C 

,     f±fVi4;                    fK3±VT3);  f9;                     r  ±  4,  ±  14  : 

*l±fVl4.                 'lKl±Vl3).  '  1 6.                 'i±l,>4 

'■0: 


.(_  3  ±  2V-  15);  6    f  ±  8,  ±  2  V=l  .  <  3,  4  ; 

1,^(_3±2a^15).  *\±2,  T8V^1.  '   U,  3. 


0    /±fV6,  ±3;         -    12,3;       1Q  f  3,  2,  J(5  ±  V-  151);        n     f  3,  1  ; 

*<-±*V6,  ±1.  '13,  2.  *  1 2,  3,  i(5TV^lM).  *  1 1,  3. 

13    f±3,  ±fV2;  14    f  ±0,  ±4;  15    /  3,  -  1 

'I  ±2,  ±$V2.  'l±4,  ±9.  11,-3. 


12   l*^*' 

14 ±»- 

Examples  CI 
1.    20,15.     2.    20,15.     3.    10,  $125.     4.   4,5.     5.   Dis.,  15  ;  rates,  3,  2£,  4. 
6.    48,  42.         7.    6,  \\.         8.    Dis.,  46$  or  30  ;  rate,  4  or  3.         9.    Rowing, 
4i  and  6  ;  walking,  \\.         10.    16,  5f 

Examples  CII 

3.  x  =  3  renders  /(a;)  a  max.  /(x)  at  a  max.  =  9. 

4.  x  =  4  renders  /(x)  a  min.  /(x)  at  a  min.  =  3. 


408  ANSWERS 

5.  x  =  —  2  renders  /(x)  a  min.  /(x)  at  a  min.  =  0. 

6.  x  =  2  renders  /(x)  a  max.  f(x)  at  a  max.  =  15. 

7.  x  =  2  renders  /(x)  a  min.  /(x)  at  a  min.  =  0. 

8.  x  =  4  renders  /(x)  a  max.  /(x)  at  a  max.  =  50. 

9.  x  =  0  renders  /(x)  a  min.  /(x)  at  a  min.  =  11. 
10.  x  =  —  £  renders  /(x)  a  max.  /(x)  at  a  max.  =  7f. 

Examples  CIII 

2.  x  =  —  1  renders  /(x)  a  min. 

3.  x  =  —  |  renders  /(x)  a  max. 

4.  x  =  4  renders  /(x)  a  max. 

5.  x  =  2  renders  /(x)  a  min. 

so 

Examples  CIV 

8;  /(x)  at  a  min.  =  —  £,  when  x  =  3.    /(x)  at  a  max.  =  —  £,  when  x  =  1. 
9.  /(x)  at  a  min.  =  b,  when  x  =  a. 

10.  /(x)  at  a  max.  =  |,  when  x  =  3. 

11.  /(x)  at  a  max.  =  ^,  when  x  =  —  5. 

12.  /(x)  at  a  min.  =  0,  when  x  =  —  1. 

13.  /(x)  at  a  max.  =  a,  when  x  =  a. 

14.  /(x)  at  a  min.  =  6,  when  x  =  ±  2.    /(x)  at  a  max.  =  10,  when  x  =  0. 

15.  /(x)  at  a  min.  ss  —  §,  when  x  =  2.    /(x)  at  a  max.  =  1,  when  x  =  0. 

16.  /(x)  at  a  max.  =  &,  when  x  =  a. 

17.  /(x)  at  a  min.  =  —27,  when  x=  ±  V2.   /(x)  at  a  max.  =  —  15,  when  x=0. 


/(x)  at  a  min. 

=  h 

/(x)  at  a  max. 

=  6. 

/(x)  at  a  max. 

=  -  2  m. 

/(x)  at  a  min. 

_  n  —  4 

{ 

fy  at  a 
lx  at  a 


Examples  CV 

y  at  a  max.  =  6,  when  x  =  1.     y  at  a  min.  =  —  2,  when  x  =  1. 
x  has  neither  a  max.  nor  a  min. 

max.  =  4,  when  x  =  2.  y  at  a  min.=  —  2,  when  x  =  2. 

max.  =  2  +  V22,  when  y  =  \.     x  at  a  min.=2—  V22,  when  y=l. 
m    f  y  at  a  max.  =  5,  when  x  =  0.  y  at  a  min.  =  —  3,  when  x  =  0. 

I  x  at  a  max.  =  4,  when  y  =  1.  x  at  a  min.  =  —  4,  when  y  =  1. 

Examples  CVI 

3.   6  and  6.    Min.  val.  =  72.      4.  A  square,  each  side  40  rd.       5.  384  sq.  yd. 
7.    6  knots,  32  m.  8.   Dimensions,  8  by  16.     Area,  128  sq.  rd. 

9.    The  corners  are  at  the  middle  points  of  the  given  square.     Side  of  min. 

square  =  — •  11.    The  length  is  twice  the  breadth. 

V2 
12.   Each  side  is  \y/2  times  the  hypotenuse.  13.    7  ft.  by  If  ft. 


ANSWERS  409 

Examples  CVII 

3.  oo.     4.  oo.     5.  0.      6.  0.     7.  &      8.  |.      9.  f.      10.    -.      11.  f.     12.  1. 

Examples  CIX 

2.   dy  =  (3  x2  +  10  as  -  6)dz.  3.  tZy  =s  -  (3  x~4  +  8  x)dx. 

4.  cty=  (28x3-f  6x-3x~*jdx.  5.  dy  =  10(as*  +  2  x"5  -  x)dx. 
7.    dy  =  120(2  +  3  x2)3xdx.  8.  dy  =  -120(1  -  5  x2)2xdx. 
9.   dy  =  5  6(o  +  bx*)%x2dx.  10.  dy  =  12(3  -  x3)-3x2dx. 

12.   dw  =  3  x2y*dx  +  $  x3y*dy.  13.  dw  =  (y2  -  3)dx  +  2  xydy. 

14.   du  =  (y3  +  2)dx  +  3(x  -  1)  y2dy.  15.  (x  +  a)$<te  +  f  x(x  +  a)  *dx. 

17.   dy=.     3<to    .  18  du  =  3x*ydx-2x*dy, 
(1  +  x)4                                                                  y« 

20.   dv  =  — ^— .  21.   dy=4V~3x±dx. 

2Vl  +  x  V2x^"3z2 

Examples  CX 

2.  ^=3x2-6x.  3.   ^=9(x3-5)2x2.  4.  /'(afl  = — . 

dx  dx       K  J  J  w     Vx2^ 

5.  /'(x)  =  2x+l-3x-*.        6.   &  = **-_.        ,    m=s__Jj2i_. 

V  ,»  dx         (1  +  x)3  '  V  '         (x  +  a)» 

9.    «J  =  2.  10.    ^  =  36  x2  -  10.  11.  /"(x)  =  450(2  + 5  x). 

dx2  dx2 

12.   12(a  +  x)-6.         13.   &y=_AH± —  14.   80(2  x2  -  3)2(14  x2  -  3). 

dx2      (a  +  x)3 

Examples  CXI 
1.    l  +  3x  +  4x2  +  7r3  +  Hx4  +  etc.      2.   2  -  5x  +  3x2  +  2x8-  5x*  +  etc. 

3.  l  +  2x  +  x2-x3-2x4-x5  +  etc.      4.    l-x-x2  +  5x3-7x4-x8  +  etc. 
5.    5  +  27  x  +  130  x2  +  623  x3  +  etc.       6.    1  +  2  x  +  3  x2  +  6  x3  +  9  x4  +  etc. 
7.   2  +  2  x  -  3  x2  -  5  x3  -  x4  -  etc.        8.    1  +  x  +  x2  -  x3  +  etc. 

9.   }  +  fx  +  3x2  +  Hz3  +  etc 

Examples  CXII 

1.  fB'Hfar^+ff+Ws  +  Hf^+etc.       2.  ar1+3+2  x-5x2-16x3+etc. 

3.  |x  +  $ x3  +  2V s6  +  ir^7  +  etc-        4-  x-2  -  aT1  -  2  x  +  2  x2  -  4  x3  +  etc. 
5.  x-x2-2x3-5x4-12x5-etc.    6.  f  x-3-^-2- £x_1 +  f|  +  $f  x  +  etc. 

Examples  CXIII 

2.  1  +  \  x  +  f  x2  -  T\  x8  +  Tf  f  x4  +  etc.         3.    1  +  ^  x  -  f  x2  +  ^  x8  +  etc. 

4.  3  +  ix-tt$a*  +  flftz*  +  etc.    6.  1  +  £x+  f  x2-  if  x8  ^^x4  +  etc. 


410  ANSWERS 

Examples  CXIV 

3.    -? I +  __1 4  3  * 

x     2(1 -x)      2(1+ a)  2(x-2)      2x 

5.—* ? 6_       5  4 


3(x-2)      3(x  +  l)  x-4     a: -3 

7    __2 5  1  8        2  23 


(x-3)3      (x-3)2     x-3  x  +  5      (x  +  5)2 

9.  -J 2 !_.  10.  »  4 


x+1      (x+1)2      (x  +  1)3                       5(5  x- 2)      5(5  x- 2)3 
11.    -2+^  +  _^_.  12.    _J *_  +         9 


x     3x+l      2x-5  2(x+l)      x  +  2      2(x  +  3) 

u.  _2  +  _^  +  ^«  u.  I  +  -L+-L  +  .     i 


x     x  +  2      (x  +  2)2  x     x  -  1      x  -  2      (x  -  2) 

Examples  CXV 
3    2x-  11  _     2  4         1  2  x  -  2 


x2  +  1        x  -  4  x2  +  1      (x2  +  1) 


x  +  1  6_  1  x  +  2 


3(x-l)      3(x2+2)  2(x-l)      2(x2+x  +  4) 

-        7  5x  -  3  8    __as 1 

x  -  1     x2  +  x  +  l"  x2  +  2     x2  +  x  +  2 

Examples  CXVI 

3.  xH|x"^t/-| x~ V  +  TV x~ V  -  ili £~V  +  etc. 

4.  x~2  +  2  x"3*/  +  3  x"4!/2  +  4  x~5yz  +  5  x~6*/4  +  etc. 

5.  x~3  -  |  %%  +  f  x~V  -  |f  x~*V  +  ^  aT^V  -  etc. 


1  _2  _5 

1  r»    3n  1  v    ~Sifi  J §_  ~    3 


V, 


6.  x3  +  }  x  3y  -  )  x  V  +  ^  x  3?/3  -  J&  x  ~3  y4  +  etc. 

7.  x^  +  f  x^y  +  f  x~V  -  TV  x-^/3  +  Tf -g  x~V  -  etc. 

Examples  CXVII 

2.  3  x5  -  2  x2  +  (15  x4  -  4  x)h  +  (30  x3  -  2)fr2  +  30  x2ft3  +  15  x/i4  +  3  h*. 

3.  2  x4  -  4  x3  +  x2  -  5  +  (8  x3  -  12  x2  +  2  x)h 

+  (12  x2  -  12  x  +  l)/i2  +  (8  x  -  4)/i3  +  2  A4. 

Examples  CXVIII 

2.  16  x4  +  32  xsy3  +  24  x2?/6  +  8  xy9  +  ?/12. 

3.  x2  -  2  x4y  +  4  x6?/2  -  8  x8#3  +  etc. 

4.  x*  -  f  x~V  _  4  ^.-1^4  _  1 2  x-fy6  _  etc. 


ANSWERS 


411 


3. 

4. 
6. 

7. 

11. 
15. 

16. 
17. 
18. 
19. 
23. 


Examples  CXIX 
a?  -  6  cfib  +  15  a462  -  20  «363  +  15  aW  -  6  a&6  +  66. 
x7  -  7  x«y  +  21  x5y2  -  35  x*y*  +  35  x3y4  -  21  x2y5  +  7  xy«  -  y7. 
1  +  4  x  +  6  x2  +  4  x3  +  x4.  5.    1  -  5  y  +  10  y2  -  10  y3  +  5  y4  -  y6. 

x-2  -  2  x_3y  +  3  x~4y2  -  4  x~6y3  -f  5  x-*y*  -  etc. 

X3         X4  X5  X6  X' 

1  -  $a2  -  | a4  -  TV«6  -  Tf^a8  _  etc. 
a6  +  16  a "^  +  96  «J^  +  256  a*  +  256  at. 


2  2 


* 


8 


+  etc. 

1+  3  x2  +  6  x4  +  10  x6  +  15  x8  +  etc. 
a^  -  \  a~^x2  -  $  af^x4  -  ^  a^x6  -  etc. 


1  x2         2x4 

27  a3      27  a4      81  a* 


10  x3 


352. 


729  a 
1  +  4  x  +'2  x2  -  8  x'' 


+  etc.         22.    2(x2  +  6  x  +  1). 

5  x4  +  8  x5  -f  2  x6  -  4  x7  +  x8. 


Examples  CXX 

=  |  log x  +  i[log  (1  +  x)  +  log (1  -  x)]. 

3.  log y  =  £[log  s  +  log  (s  -  a)  -  log  6  -  log  c]. 

4.  logy  =  £[loga  +  21ogfi  -f  41ogc  —  51ogd]. 

5.  logy  =  2  log  a  +  Klog(«  +  &)+  log(a  -  6)-  log(l  +  &)]. 

6.  logy  =  £[logx-f  log(l-x)]-£logz. 

7.  logy  =  £[raloga  +  plogfo  -  «logc]. 

8.  logy  =  $[log(a+  b)  +  log  (a  -  b)-  log  a]. 

9.  log y  =  i [log x  +  log  (x  -  4)  -  log  (x  +  1)]. 


11. 
15. 
18. 
21. 


,„ ,     mdx 

ay  = 

x 


dy  = 


dy 


x  =  — 


_  2  mxdx 
a2  -  x2' 
log  ft 
log  a 

x  =  2£,  y  =  |. 


2.902003. 
1.909930. 
2.888089. 


mdx 
1-x 

16.'  dy  = 
19.   x 


3  mdx 


13. 

x 

4  mxdx 
1  +  x2' 
log  a  —  log  b 
clog  a 


14.   dy  = 
17.  x 


mdx 

x 
logc 


2  V  log  a      logbl 

Examples  CXXI 

2.    0.935507.  3.    3.891203. 

6.   3.930236.  7.    4.896273. 

10.   3.938648.  11.    1.895266. 


b  log  a 
x  =  log  7. 


4.    1.915716. 

8.    1.929092. 

12.    3.934229. 


412 


ANSWERS 


Examples  CXXII 

1.    8313.                       2.    792.75.  3.    8.1743.  4.  86764.6. 

5.    79.8252.                  6.    .83752.  7.    .0773742.  8.  .0085167. 

9.  860900.       10.  820000.  11.  .008.  12.  819554. 

13.  8636.32.  14.  85.8052.  15.  790.3972.  16.  7.75987. 
17.  5.4283.       18.  5.4641. 


Examples  CXXIII 

5.    1,  1.         6.    7,  1  ;  2,  4.         7.    5,  1  ;  19,  6  ;  33,  11  ;  etc.         8.    1,2;  3,  1. 

9.    Impossible.  10.    20,  3  ;  39,  7  ;  etc.  11.    17,  1 ;  10,  4  ;  3,  7. 

12.    1,  53  ;  3,  40  ;  5,  27  ;  7,  14  ;  9,  1.  13.    3,  7  ;  8,  21  ;  13,  35  ;  etc. 

14.    Impossible.  15.   2,  1,  3.  16.  4,  2,  7. 

Examples  CXXIV 

1.    16,  2  ;  12,  5  ;  8,  8  ;  4,  11.  2.    11,  15.  3.    4  ways. 

4.    2,  74;  15,  55;  28,  36;  41,  17.      5.    Impossible.  48  ways.       6.   f,  £,  f,  etc. 

7.    6,3.        8.    117,  19;  77,  59;  37,  99.        9.    5,1.         10.    A  gives  £6  and 

receives  $28.        11.    1,  6,  8,  or  3,  3,  9.        12.   5,  10,  15. 


3.   y*  +  3y*  +  486  =  0. 


Examples  CXXV 

6.    y5-2y*-2ys  +  45y*-  12*/ +  81  =  0. 


2.    19. 


Examples  CXXVI 
3.    1.  5.    7.  6.    12. 


7.    0. 


Examples  CXXVII 

2.  1  pos.,  1  neg.,  2  imag.  3.    No  pos.,  1  neg.,  4  imag. 

4.  1  pos.,  no  neg.,  4  imag. 

5.  Not  more  than  2  pos.,  not  more  than  2  neg.,  at  least  2  imag. 

6.  Not  more  than  3  pos. ,  not  more  than  1  neg. ,  at  least  2  imag. 

7.  1  pos.,  not  more  than  3  neg.,  at  least  2  imag. 

8.  Not  more  than  3  pos.,  1  neg.,  at  least  2  imag. 

9.  1  is  the  only  real  root.  10.    1  and  -  1  are  the  only  real  roots. 
11.  —  1  is  the  only  real  root.                 12.    No  real  root. 


2.   2,  3,  -  4. 

14.   2,  ±V3,  ±VS. 


Examples  CXXVIII 

5.    7,  ±V5.  9 

17.    1,  1,  -  1,  |,  f.  20 


1,1,  -1,  i(_3±V5). 
1,  1,  1,  2,  2,  -  2,  -  3. 


AN S WEBS  413 

Examples  CXXIX 

4.    ±  V2,  ±  V2,  ±  VS.  8.    ±  2,  ±  V£,  ±  V6. 

12.    ±  V2,  ±  V6,  ±  V^T,  ±  a/^2.        13.    0,  ±  V2,  ±  V3,  ±  V6,   ±  V^2. 
14.    +_  V^l,  ±  \^~2,  ±  V^3,  ±  V^5.  17.    3,  ±  V2,  ±  V3,  ±  V5. 

Examples  CXXX 

7.    4x5  +  24  x4  +  15x3  -  80  x2  -  39 x  +  36  =  0. 
9.   «6-G^  +  8x*  +  18x3-  63s2  +  24* +  20  =  0. 

Examples  CXXXI 

4.  Between  0  and  1,  0  and  —1,-5  and  —  6. 

7.   Between  3  and  4.    Two  imag.        14.  Between  0  and  1, 1  and  2,  2  and  3. 
18.    Between  3.2  and  3.3,  3.4  and  3.5,  -  1  and  -2,-3  and  -  4. 
22.    Between  1  and  2,  0  and  —1,-1  and  —  2.    Two  imag. 

Examples  CXXXII 

2.  2  +  V3  and  2  —  V3  are  each  double  roots.  4.   3,  1  ±  VE1  1  ±  V5. 
•. ±  V§,  ±  V2f  1  ±  VS.                                     9.    ±  V2,  ±  V2,  ±  V2,  ±  V3. 

10.  1  +  VS  and  (1  —  V3)  are  each  quadruple  roots. 

Examples  CXXXIII 

3.  x3  +  20.630  x2  -  . 343893  x  +  .000043597  =  0. 

4.  x4  +  11.4x3  +  36.735  x2  +  36.1965  x  -  .29499375  =  0. 

Examples  CXXXIV 

3.  2.7824,  4.166,  -  .9489.  4.   4.117.    Two  roots  imag. 

5.  4|,  2  ±  VS  ;  or  4.125,  .26795,  3.73205.      6.    -  2.2134.    Two  roots  imag. 
7.    9.9666.     Two  roots  imag.  8.    2.25,  3.6055,  -  3.6055. 

9.    -2.1768.    Two  roots  imag.                    10.  2.3569,2.6920,-2.0489. 

11.  2.4641,  -  4.4641.     Two  roots  imag. 

12.  4.3166,  -2.3166.    Two  roots  imag.        13.  .337,  1.636,  -  1.691,  -  4.283. 
14.   Two  roots  imag.                                       15.  4.5814.    Four  roots  imag. 
16.  4.5195.    Four  roots  imag.         17.    5.874.  18.    .4148.         19.    106.474. 

Examples  CXXXVI 

4.  Between  0  and  —1,-2  and  —  3,  two  between  2  and  3. 

6.  Between  —  1  and  —  2,  4  and  5,  three  between  1  and  2. 

7.  Between  —  1  and  —  2,  4  and  5,  two  between  1  and  2,  two  imag. 


414 


ANSWERS 


11.    Double  root  between  -1  and  -2,  double  root  between  1  and  2,  two  imag. 
13.   Between  —  1  and  —  2,  3  and  4,  double  root  between  —  1  and  —  2,  double 
root  between  1  and  2. 


Examples  CXXXVII 
2.    1,  1,  |(1  ±  V^~3).      3.   2  ±  V3,  1(1  ±  V="3). 
5.    ±1,  K-7±3V5). 
7.    ±1,1(1±VU72).  8.    1,3,1,-2,-1. 

10.    2,  i,  2,  |,  K1±V^3). 


4.    ±1,  i(l±V^3). 
6.    |(3±VS),  ^(-7±3V5). 


9.    1,  ±V-1,  i(-l±V-35) 
2.    1,  i(-l±V^3). 


5.    ±V±v^T. 

7.   3  times  the  roots  of  ex.  1. 

9.    ± 


Examples  CXXXVin 

3.    -1,  i(l±v^3).  4.    ±1,  ±V=1. 

6.    -  1,  i(l  ±V5±V- 10±2V5). 


8.    ±1,  ±Vi(-l±V-3). 


1,  ±Vki±V-3). 
1.   Convergent. 


Examples  CXXXIX 
3.    Convergent. 


4.    Divergent. 


2.    Convergent. 
6.    Convergent. 


Examples  CXL 

1.   Convergent  when  x  <  1,  divergent  when  x>l. 

3.    Convergent  when  x  <1.  4.    Convergent. 

7.   Divergent.  8.    Convergent  for  x  >  1,  divergent  for  x<  1. 

9.    Divergent.  10.    Convergent. 

11.   Convergent  for  x<  1,  divergent  for  x  >  1.  14.   Convergent. 

15.   Convergent  when  X  is  numerically  equal  to  or  less  than  1,  divergent 

when  x  is  numerically  greater  than  1. 


16.   Convergent. 

2.    1,6;  408  x5. 
5.    1,  1,  -1;  4x7. 

2-    0. 


17.    Convergent. 

Examples  CXLI 
3.    2,  1  ;  99  x5. 
6.    -  1,  -1  ;  2x6. 

Examples  CXLII 
3.    0,  0. 


18.    Divergent. 

4.    -  2,  3  ;  489  x5. 
7.    2,  1,  -2;  173  x\ 

4.    8,  32. 


ANSWERS  415 

Examples  CXL.III 
2.    78.  3.   225.  4.   6396  a;". 

Examples  CXLIV 

10   &=5(§-(.  +  iK»  +  2))i  s«  =  i 

11.    «,  =  ___________;    «.  =  -• 

i  o      o    3  2        , 1 t      n    o 

13     An_I      rc  +  2  +  2(N  +  l)(rc  +  2)'    ^_4* 

n~6      7i  +  3~f"(tt  +  3)0  +  4)'       °°~6 


15. 


n"2V        l-3.5...(2«  +  l)j'      ac~2* 
Examples  CXLV 


J? 

1  + 11 4 

3 

1  +  a;              4      1  +  3  3           -           1+3 
1  -  X  _  3*                   (1  _  a;)2              '1-23-2  X* 

1  +  2  a;  -  3  32 

6. 

2  +  5a; 

7          1+3                                   g              3-3- 6  32 

(1  -  3)2                                       1  -  2  3  -  32  +  2  33 

1  +  5  x  +  4  x2 

9. 

1  +  2  36  -  3  X2 
1  -  *  +  X'2  +  Xs 

10-         1"-6x  +  8a!'       ;p  =  2,  g  =  3,  r  =  4. 
1-23-3  32  -4  33 '  ^           y 

Examples  CXLVI 
2.    400.  3.   2548.  4.    1275.  5.    278,256.  6.    147. 

Examples  CXLVII 

1.  680.       2.  1540.       3.  1505.       4.  624.       5.  1240. 
6.  2730  7.  7490.  8.  3880.  9.  36,256. 

Examples  CXLVIII 

2.  10  hr.  19  min.  47.45  sec.  3.    10  hr.  6  rain.  39.1  sec. 
4.    3.97905.                                                 5.   44°  58  40'- 


416        .  ANSWERS 

Examples  CXLIX 

1.    336.  2.    60.  3.    72,  504,  3024,  15,120,  60,480,  181,440,  362,880, 

362,880.           4.    720.           5.    210.           6.    360.            7.    840.  8.    40,320. 

9.    360.        10.  2520,  4,989,600.        11.  64.        12.  623,529.        13.  7,257,600. 

14.    (a)  5040,    (6)  [7  x  |8  x  [5  x  |3.                    15.    1260.  16.    5040. 

17.    (a)  576,  (6)  144.          18.    576.           19.    120.           20.    720.  21.    120. 

22.  2880.  23.  181,440.  24.  408,408.  25.  4.  26.  6.  27.  8. 
28.   15.            29.    7.            30.    3.            31.    13. 

Examples  CL 

1.    15.         2.    10.         3.   36,  84,  126,  126,  84,  36,  9,  1.         4.    15.  5.    792. 

6.  15,890,700.    7.  210.    8.  45.    9.  Loses,  $4.90.  10.  1716. 

11.  31.    12.  26.    13.  4095.    14.  63 ;  yes.    15.  3255.  16.  322. 

17.  35.      18.  129.      19.  479.      20.  2520.  21.  1980. 

152         152 
22.  (a)   -*=-  ;  (b)  — 23.  945. 

V  ;  ({Uy  '  W  [4  x  ([13)4 


14  DAY  USE 

RETURN  TO  DESK  FROM  WHICH  BORROWED 

LOAN  DEPT. 

This  book  is  due  on  the  last  date  stamped  below,  or 

on  the  date  to  which  renewed. 

Renewed  books  are  subject  to  immediate  recall. 


4V'57Pft 


ma:;  a  oi35 

LIBRARY  HSR 


SEP1  61969 


RECDLD 
SEP  16  1259 


TW 


t&&L 


rtcOD  LD 


DE0  2  9'64 


U& 


LD  21-100m-6,'56 
(B9311sl0)476 


General  Library 

University  of  California 

Berkeley 


r 


911285     „  J, 

6}  A  IS  3    ■ 


THE  UNIVERSITY  OF  CALIFORNIA  LIBRARY 


